Nilpotent Quotient Group (6.1.6)

Dummit and Foote Abstract Algebra, section 6.1, exercise 6:

If $G/Z(G)$ is nilpotent, prove $G$ is nilpotent.

Proof: Lemma 1: For $φ$ an isomorphism, we have $φ(Z(G))=Z(φ(G))$. Proof: For any $x,y∈G$, we observe $$φ(x)∈φ(Z(G))⇔x∈Z(G)⇔xy=yx⇔φ(xy)=φ(yx)⇔$$$$φ(x)φ(y)=φ(y)φ(x)⇔φ(x)∈Z(φ(G))~~~\square$$

Now, let the overbar denote passage into $G/Z(G)$.

Inductively prove that $Z_{n}(\overline{G})=\overline{Z_{n+1}(G)}$ for all nonnegative $n$. When $n=0$, we have $Z_0(\overline{G})=Z(G)/Z(G)=Z_1(G)/Z(G)=\overline{Z_1(G)}$, so now validate the inductive step: Let $φ : G/Z_n(G)→\overline{G}/\overline{Z_n(G)}$ be an isomorphism by $φ(gZ_n(G))=\overline{g}\overline{Z_n(G)}$ (the inverse of the natural isomorphism by the Third Isomorphism Theorem), and we have:
$Z_n(\overline{G})/Z_{n-1}(\overline{G})=Z(\overline{G}/Z_{n-1}(\overline{G}))⇒$
$Z_n(\overline{G})/\overline{Z_n(G)}=Z(\overline{G}/\overline{Z_n(G)})⇒$
$Z_n(\overline{G})/\overline{Z_n(G)}=Z(φ(G/Z_n(G)))⇒$
$Z_n(\overline{G})/\overline{Z_n(G)}=φ(Z(G/Z_n(G)))⇒$$Z_n(\overline{G})/\overline{Z_n(G)}=φ(Z_{n+1}(G)/Z_n(G))⇒$
$Z_n(\overline{G})/\overline{Z_n(G)}=\overline{Z_{n+1}(G)}/\overline{Z_n(G)}$.
Since $Z_{n+1}(G)$ contains $Z_{n}(G)$ contains $Z(G)$, we have $\overline{Z_{n+1}(G)}$ contains $\overline{Z_n(G)}$. Further, $Z_n(\overline{G})$ contains $\overline{Z_n(G)}=Z_{n-1}(\overline{G})$. The Lattice Isomorphism Theorem describes a bijection between the subgroups of a parent group containing the normal subgroup and the subgroups of the quotient group, so we are left with $Z_n(\overline{G})=\overline{Z_{n+1}(G)}$.

Now, set $n=m$ where $Z_m(\overline{G})=\overline{G}$ and we have $Z_{m}(\overline{G})=\overline{Z_{m+1}(G)}=\overline{G}$, so similarly as above $Z_{m+1}(G)=G$, therefore $G$ is nilpotent (and $\overline{Z_{m}(G)}=Z_{m-1}(\overline{G})≠\overline{G}$ so $Z_{m}(G)≠G$, therefore $G$ is of class $m+1$ when $G$ is nontrivial). $\square$

Normal Subgroups and Center of a Nilpotent Group (6.1.2a)

Dummit and Foote Abstract Algebra, section 6.1, exercise 2a:

Let $G$ be finite and nilpotent, and $1 < H \trianglelefteq G$. Prove $H \cap Z(G) ≠ 1$.

Proof: Let $|G| = p_1^{\alpha_1}...p_r^{\alpha_r}$.

By theorem 3(4), we have $G ≅ P_1 \times ... \times P_r$. Associate $H$ under this isomorphism, and for some nontrivial element in $H$ choose a coordinate (the $i^{th}$) in which a nontrivial element resides; call the associated Sylow $p_i$-subgroup $P_i$. Due to the way by which direct products work, for any $x∈A$ we are able to identify some element in $H$ with $x$ in the $i^{th}$ coordinate (*), where $A$ is the group generated by the $i^{th}$ coordinates of the elements of $H$. Since $H \trianglelefteq G$, we must have $A \trianglelefteq P_i$, since an arbitrary element of $P_i$ associated within $G$ may be used to conjugate the elements of $H$; if one of these conjugations results in an element in the $i^{th}$ coordinate not contained in $A$, then since $H$ is normal, this has contradicted the construction of $A$. Therefore, by theorem 1, choose a nontrivial element of $A \cap Z(P_i)$ and associate it with an element in $H$ (*), and raise this to the power of $|G|/p_i^{\alpha_i}$. This returns an element of $H$ that has the identity in every coordinate besides the $i^{th}$, and in the $i^{th}$ an element within $Z(P_i)$ that is nontrivial by Lagrange. Therefore, this nontrivial element is within $Z(G)≅Z(P_1) \times ... \times Z(P_r)$ by 5.1.1. $\square$

Characteristicity of the Central Upper Series (6.1.1)

Dummit and Foote Abstract Algebra, section 6.1, exercise 1:

Prove $Z_i(G)~\text{char}~G$ for all $i$.

Proof: Lemma 1: $Z(G)~\text{char}~G$ for any group $G$. Proof: For any $x,y∈G,~φ∈Aut(G)$ we have:$$φ(x)∈Z(G)⇔φ(x)φ(y)=φ(x)φ(y)$$$$⇔φ(xy)=φ(yx)⇔xy=yx⇔x∈Z(G)~~~\square$$
Lemma 2: For $K~\text{char}~G$ and $φ∈Aut(G)$ and letting the bar notation denote passage into $G/K$, we have $ψ∈Aut(\bar{G})$ where $ψ: \bar{x} \mapsto \overline{φ(x)}$. Proof:
Well-defined: $\bar{x}=\bar{y}⇒y^{-1}x∈K⇒φ(y^{-1}x)∈K⇒\overline{φ(x)}=\overline{φ(y)}⇒ψ(\bar{x})=ψ(\bar{y})$.
Homomorphic: $ψ(\overline{xy})=\overline{φ(xy)}=\overline{φ(x)φ(y)}=\overline{φ(x)}~\overline{φ(y)}=ψ(\overline{x})ψ(\overline{y})$.
Injective: $$ψ(\overline{x})=ψ(\overline{y})⇒\overline{φ(x)}=\overline{φ(y)}⇒\overline{φ(y^{-1}x)}=1$$$$⇒φ(y^{-1}x)∈K⇒y^{-1}x∈K⇒\overline{x}=\overline{y}$$.
Surjective: For any $\overline{x}$, let $y$ be the preimage of $x$ by $φ$. By way of construction, $ψ(\overline{y})=\overline{x}$.$\square$

Returning to the main result, we shall proceed by induction. Clearly $Z_0(G)~\text{char}~G$, so apply the inductive hypothesis on $i$. Let the overbar denote passage into $G/Z_{i-1}$. Since $\overline{Z_i(G)}=Z(\overline{G})$, by lemma 1 we have $\overline{Z_i(G)}~\text{char}~\overline{G}$. Let $φ$ be any automorphism of $G$, and let $ψ$ be the automorphism of $\overline{G}$ afforded by $φ$. Complete the proof: $x∈Z_i(G)⇔\overline{x}∈\overline{Z_i(G)}⇔ψ(\overline{x})∈\overline{Z_i(G)}⇔\overline{φ(x)}∈\overline{Z_i(G)}⇔φ(x)∈Z_i(G)$. $\square$

Heisenberg Group Over Prime Fields (5.5.25)

Dummit and Foote Abstract Algebra, section 5.5, exercise 25:

Prove that $H(\mathbb{F}_2)≅D_8$ and for $p$ an odd prime prove that $H(\mathbb{F}_p)$ is of exponent $p$ and isomorphic to the first nonabelian group presented in example 7: $$\langle x,a,b~|~x^p=a^p=b^p=1,~ab=ba,~xa=abx,~xb=bx \rangle$$

Proof: One can easily check that the groups generated by $\left( \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right)$ and $\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right)$ are of orders 4 and 2 and have trivial intersection and thus generate $H(\mathbb{F}_2)$ of order 8, and indeed the generators fulfill the qualities of $r$ and $s$ in the presentation of $H(\mathbb{F}_2)$.

We can see $\left( \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$ and $\left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right)$ don't commute in any Heisenberg group over a prime field, so that by previous investigations of groups of order $p^3$ the center of $H(\mathbb{F}_p)$ is of order $p$. Now, $Z=\langle \left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) \rangle$ is of order $p$ and is clearly within (and thus completely comprises) the center. Prove using induction that $\left( \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)^n=\left( \begin{array}{ccc} 1 & n & n \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$ and that $\left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right)^n=\left( \begin{array}{ccc} 1 & 0 & n \\ 0 & 1 & n \\ 0 & 0 & 1 \end{array} \right)$ for all $n∈\mathbb{Z}^+$, and let $A=\langle \left( \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) \rangle$ and $B=\langle \left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right) \rangle$ be subgroups of order $p$. Due to the normality of $Z$ we can tell $AZ$ is a group, and due to its index it is normal in $H(\mathbb{F}_p)$. Now, $B∩AZ = 1$ and $BAZ=H(\mathbb{F}_p)$ so that $H(\mathbb{F}_p)=AZ \rtimes_φ B$ for some homomorphism $φ : B → Aut(AZ)$. Since $Z=Z(H(\mathbb{F}_p))$, we know that the exponent of $AZ$ is $p$, so that $AZ≅Z_p \times Z_p$ and thus $|Aut(AZ)|=p(p-1)^2(p+1)$, and in particular the Sylow p-subgroups of $Aut(AZ)$ are of order $p$ and already established to be conjugate, so that by exercise 6 there is only one unique nonabelian construction of $AZ \rtimes_φ B$, which must be $H(\mathbb{F}_p)$. Mapping the generator of $B$ to the element of $GL_2(\mathbb{F}_p)≅Aut(AZ)$ described in the example, we obtain the presentation mentioned (with the generator of $B$ in place of $x$, and the generator of $Z$ in place of $b$, and the generator of $A$ in place of $a$).

As for the exponent, under their notation we can see $b∈Z(G)$, leading to $(x^{n_1}a^{n_2}b^{n_3})^p=(x^{n_1}a^{n_2})^p$. Migrating the $a$ terms to the left of the block using the quality $xa=abx$ and then migrating the $b$ terms similarly since $b$ is in the center, we obtain $(x^{n_1}a^{n_2})^p=a^{pn_2}b^{m}x^{pn_1}$, where $m$ is determined by the process. Note that we migrate the first $a^{n_2}$ term through 1 block of $x^{n_2}$'s to obtain $n_1n_2$ resultant $b$'s, and we migrate the second $a^{n_2}$ term through 2 blocks of $x^{n_2}$'s to obtain $2n_1n_2$ resultant $b$'s, and so forth. Counting up all of these, we obtain $n_1n_2+2n_1n_2+...+pn_1n_2=n_1n_2(1+2+...+p)=n_1n_2 \dfrac{p(p+1)}{2}$ resultant $b$'s after the whole process, which is divisible by $p$ - the order of $b$ - when $p$ is an odd prime. $\square$

Abelian Classification Theorem (5.5.24)

Dummit and Foote Abstract Algebra, section 5.5, exercise 24:

Let $p_1^{\alpha_1}p_2^{\alpha_2}...p_r^{\alpha_r}$ be the prime decomposition of $n$. Prove: $(|G|=n ⇒ G~\text{is abelian})⇔∀i,j∈\{1,...,r\}(\alpha_i∈\{1,2\} \land p_i \nmid p_j^{\alpha_j}-1)$.

Proof: Recall that a direct product is abelian if and only if all of its factors are abelian (5.1.1).

($⇒$) (1) Assume there is a prime factor $p_k^{\alpha_k}$ in the prime decomposition of $n$ such that $\alpha_k \geq 3$. Obtain a nonabelian group $P$ of order $p_j^3$ as described in the material and we have $P \times Z_{n/p_j^3}$ is a nonabelian group of order $n$, contrary to assumption.

(2) Assume $p_i \mid p_j^{\alpha_j}-1$ for some $i,j$. If $\alpha_j = 1$, then obtain the nonabelian group $P$ of order $p_ip_j$ also as presented in the material and we have $P \times Z_{n/(p_ip_j)}$ is a nonabelian group of order $n$, also contrary to assumption. If $\alpha_j=2$, then we have $p_i \mid p_j^2-1=(p_j-1)(p_j+1)$. Now, $|Aut(Z_{p_ j} \times Z_{p_j})|=p_j(p_j-1)^2(p_j+1)$, so it contains a cyclic group of order $p_i$ by Cauchy, to which $φ$ shall nontrivially map $Z_{p_i}$. We obtain $H = (Z_{p_j} \times Z_{p_j}) \rtimes_φ Z_{p_i}$ is a nonabelian group of order $p_ip_j^2$, so that $H \times Z_{n/(p_ip_j^2)}$ is a nonabelian group of order $n$, a contradiction.

($\Leftarrow$) Assume $G$ is a nonabelian group of order $n$. Inductively, every proper subgroup of $G$ is abelian (since any proper subgroup divides $G$'s order, and any of the factors whose exponents are reduced from 2 to 1 in the process are insured since $p \nmid q^2 - 1 = (q-1)(q+1) ⇒ p \nmid q-1$), so that by 4.5.56 $G$ is solvable. Let $N$ be the normal subgroup such that $G/N$ is abelian, and in particular $N$ is not trivial due to the nonabelianness of $G$. If $N$ has a prime decompositional width greater than 1 (i.e. is divisible by two distinct primes), then two distinct Sylow subgroups $A$ and $B$ of $N$ are normal in $N$ due to $N$'s proper-subgroup abelianness, and are thus characteristic in $N$, leading to them being normal in $G$. Since $G/A$ and $G/B$ are of smaller order than $G$ and holding to the a priori qualities of $G$, and thus abelian, we have by 5.4.15 that $G/(A∩B)=G/1$ is abelian, so $N$ is of width 1, and of order $p_i$ or $p_i^2$.

If $G$ is of width 1, then by the abelianness of groups of order $p$ or $p^2$ we know that $G$ is abelian, so assume $G$ is of width greater than 1. By Cauchy, there is a subgroup $\bar{H} \trianglelefteq \bar{G}$ (normal since the latter is abelian) with an order of $p_k$ - a prime distinct from $p_i$ - so that $H \trianglelefteq G$ is of order $p_ip_k$ or $p_i^2p_k$. If $G = H$ of order $p_ip_k$, then $G$ is abelian by previous material. If $G=H$ of order $p_i^2p_k$, then since $p_i \nmid p_k - 1$ we have $p_k \not ≡ 1 \mod{p_i}$, leading to the Sylow $p_i$-subgroup $C$ being normal so that $G = C \rtimes D$ for a $p_k$-subgroup $D$, but $Aut(C)$ has order either $p_i(p_i-1)^2(p_i+1)$ or $p_i(p_i-1)$ and $p_k \nmid p_i^2-1 = (p_i-1)(p_i+1)$, so we have it is the trivial homomorphism in the semidirect product and $G$ is abelian. Therefore $H$ is a proper subgroup and thus abelian so that its Sylow $p_k$-subgroup $P_k$ is characteristic in $H$ and thus normal in $G$, so that $G/P_k$ is also abelian and now $G/(N∩P_k)=G/1≅G$ is abelian. $\square$

Routine Application of Semidirect Products (5.5.7)

Dummit and Foote Abstract Algebra, section 5.5, exercise 7:

This exercise describes thirteen isomorphism types of groups of order 56.
(a) Prove that there are three abelian groups of order 56.
(b) Prove that ever group of order 56 has either a normal Sylow 2-subgroup or a normal Sylow 7-subgroup.
(c) Assume there is a normal Sylow 7-subgroup, and construct the following groups with the Sylow 2-subgroup specified:
$~~~~~$one group when $S ≅ Z_2 \times Z_2 \times Z_2$,
$~~~~~$two nonisomorphic groups when $S ≅ Z_4 \times Z_2$,
$~~~~~$one group when $S ≅ Z_8$,
$~~~~~$two nonisomorphic groups when $S ≅ Q_8$,
$~~~~~$three nonisomorphic groups when $S ≅ D_8$.
[For a particular $S$, two groups are not isomorphic if the kernels of the maps from $S$ to $Aut(Z_7)$ are not isomorphic.]
(d) Let $G$ be a group of order 56 with a nonnormal Sylow 7-subgroup. Prove that if $S$ is the Sylow 2-subgroup of $G$, then $S ≅ Z_2 \times Z_2 \times Z_2$.
(e) Prove that there is a unique group of order 56 with a nonnormal Sylow 7-subgroup. [For existence, use the fact that $|GL_3(\mathbb{F}_2)|=168$; for uniqueness use exercise 6.]

Proof: (a) By partitioning 56 by its elementary divisors, we easily obtain the groups $Z_8 \times Z_7$, $Z_4 \times Z_2 \times Z_7$, and $Z_2 \times Z_2 \times Z_2 \times Z_7$.

(b) If both were nonnormal, then by Sylow's theorems, we could only have $n_2 = 7$ and $n_7 = 8$. Add up the nonidentity elements of the Sylow 7-subgroups, all the elements of a Sylow-2 subgroup, and at least one distinct element of another Sylow 2-subgroup to obtain $8*6 + 8 + 1 = 57 > 56 = |G|$.

(c) For the first three cases, note that a homomorphism defined on an abelian group's factor's generators extends uniquely to a group homomorphism since up to multiplicative rearrangement of the terms, there is a unique way of writing the elements of the group. This covers the gist of the proof, but further details may be obtained here as it applies to elementary abelian groups. In any case, this allows us only to define a homomorphism merely by defining it for the generators, and thus allows us to define $S$'s potential actions on $H$ as part of the semidirect product by defining the action for the generators of $S$.

Since the Sylow 7-subgroup $H$ is cyclic, we have $Aut(H) ≅ Z_6$ by its abelianness and order, so that the only nontrivial homomorphism from $S$ of order 8 into $Aut(H)$ has for image the unique subgroup of order 2, which contains the identity automorphism and the inverting automorphism. Thus, we might have $s \cdot h = h^{-1}$ for all $h∈H$ and for $s∈\{(0,0,1),(0,1,0),(1,0,0)\}$.

For the next case, it is similar, as the only available homomorphisms are those mapping $S$ to the inversion subgroup of $Aut(H)$. Define two actions of $S$ on $H$ thus: $s \cdot h = h^{-1}$ for $s∈\{(1,0),(0,1)\}$, or $s \cdot h = h^{-1}$ for $s∈\{(1,0)\}$ and $s \cdot h = h$ for $s∈\{(0,1)\}$. The former utilizes a homomorphism from $S$ into $Aut(H)$ with a kernel of order 4 and containing $(1,1)$ of order 4, so the kernel is isomorphic to $Z_4$. The latter produces a homomorphism with kernel of order 4 containing two normal, trivially intersecting subgroups of order 2 $\langle (2,0) \rangle$ and $\langle (0,1) \rangle$ so that it is isomorphic to $Z_2 \times Z_2$. Thus these two semidirect products afforded by these actions are distinct by the footnote.

For the third case, we must have $s \cdot h = h^{-1}$ for $s∈\{x\}$ where $x$ generates $S$.

For the fourth case, we can have $s \cdot h = h^{-1}$ for $s∈\{i,j\}$, and compare it with the trivial homomorphism producing $H \times Q_8$. The former has a nontrivial kernel, so these two are distinct.

For the fifth case, we can have $s \cdot h = h^{-1}$ for $s∈\{r,s\}$, and $s \cdot h = h^{-1}$ for $s∈\{s\}$ and $s \cdot h = h$ for $\{r\}$, as well as the trivial homomorphism producing $H \times D_8$. The first has a kernel of $\langle r^2, sr \rangle$, the second $\langle r \rangle$, and the third the identity, which are easily checked to be isomorphic to $Z_2 \times Z_2$, $Z_4$, and the identity group respectively, so that these are all distinct semidirect products.

(d) We have $G ≅ S \rtimes_φ H$ as usual, where $H$ is the nonnormal Sylow-7 subgroup and $S$ must thus be normal. Since $H$ is nonnormal, the homomorphism $φ$ must not be trivial, so that the kernel of $H$ is the identity. Let $H$ act on $S$ by conjugation: If $h^a \cdot s = h^b \cdot s$ for any nonidentity $h$ and $s$, then $h^a = h^b$, so that every element of $S$ can be expressed as $h^ash^{-a}$ for some integer $a$ producing a distinct element of $H$. These all have the same order as $s$, so that all of the elements of $S$ have the same order, and by Cauchy's theorem there exists an element of order 2 so that $S$ must be isomorphic to $Z_2 \times Z_2 \times Z_2$.

(e) Since $|Aut(S)| = |GL_3(\mathbb{F}_{19})| = 168 = 2^3*3*7$, we have that all the possible nontrivial homomorphisms from $H$ into $Aut(S)$ are onto the Sylow 7-subgroups thereof (map generator to generator for homomorphism), which are established to be conjugate. By this fact, the cyclicity of $H$, and exercise 6, we have the sole semidirect product obtained. $\square$

Commutators and Groups of Prime-Cubed Order (5.4.7-9)

Dummit and Foote Abstract Algebra, section 5.4, exercises 7-9:

7. Let $p$ be prime and let $P$ be a non-abelian group of order $p^3$. Prove $P'=Z(P)$.

8. Let $x,y∈G$ commute with $[x,y]$. Prove that for all $n∈\mathbb{Z}^+$, we have $(xy)^n = x^ny^n[y,x]^{n(n-1)/2}$.

9. Let $p$ be an odd prime, $φ$ be the $p$-power map $φ : x \mapsto x^p$, and $P$ be a group of order $p^3$. Prove $φ$ is a homomorphism and $\text{im}~φ ≤ Z(P)$. If $P$ is not cyclic, show that $|\text{ker}~φ|∈\{p^2,p^3\}$. Is $φ$ a homomorphism in non-abelian groups of order 8? Where is the oddness of $p$ invoked in the above proof? [Use exercise 7.]

Proof: (7) By the class equation we have $|Z(P)|≠1$ and by the non-abelianness of $P$ we have $|Z(P)|≠p^3$. Assuming $|Z(P)|=p^2$ leads to $P/Z(P)$ being cyclic and thus $P$ being abelian, so therefore $|Z(P)|=p$. By previous investigations and/or similar logic, all groups of order $p^2$ are abelian, so that $P/Z(P)$ is abelian and thus $P'≤Z(P)$. This entails either $P'$ is trivial or $P'=Z(P)$; the former implies $P/1≅P$ is abelian, so the latter is the only available possibility.

(8) First note that if $x[x,y]x^{-1}=[x,y]$, then $(x[x,y]x^{-1})^{-1}=[x,y]^{-1}$, so that by exercise 1 the right-hand side is $[y,x]$ and the left-hand side is $(x[x,y]x^{-1})^{-1}=x[x,y]^{-1}x^{-1}=x[y,x]x^{-1}$, so that $x,y$ in this exercise might be assumed to commute with $[y,x]$ as well.

Proceed by induction. The basis case $(xy)^1=xy=x^1y^1[y,x]^{(1)(0)/2}$ holds true, so assume $(xy)^k=x^ky^k[y,x]^{k(k-1)/2}$ for all $k < n$. We have $(xy)^n=(xy)^{n-1}xy=x^{n-1}y^{n-1}[y,x]^{(n-1)(n-2)/2}xy=x^{n-1}y^{n-1}xy[y,x]^{(n-1)(n-2)/2}$$=x^{n-1}y^{n-1}xy[x,y]^{n-1}[y,x]^{n-1}[y,x]^{(n-1)(n-2)/2}=x^{n-1}y^{n-1}xy[x,y]^{n-1}[y,x]^{n(n-1)/2}$. This implies that the truth of the original proposition hinges on the validity of the new proposition, $x^{n-1}y^{n-1}xy[x,y]^{n-1}=x^ny^n$. Rework this to $xy^n=y^{n-1}xy[x,y]^{n-1}$, and proceed with further induction. The basis case with $n=1$ clearly holds, so apply the inductive hypothesis and finish the problem: $xy^n=xy^{n-1}y=y^{n-2}xy[x,y]^{n-2}y$$=y^{n-2}xy[y,x][x,y]^{n-1}y=y^{n-2}xy[y,x]y[x,y]^{n-1}=y^{n-1}xy[x,y]^{n-1}$.

(9) Assume $P$ is abelian. Then $φ(xy) = (xy)^p = x^py^p = φ(x)φ(y)$ is a homomorphism and since $Z(P) = P$ we clearly have its image contained therein. Therefore assume $P$ is non-abelian. Since by exercise 7, we have $P'=Z(P)$, the setting verifies the criterion for exercise 8 that any $x,y∈P$ commute with $[x,y]∈P'$. Since $p$ is odd, we have $2 \mid p-1$, so that $p \mid p(p-1)/2$. This implies $[y,x]^{p(p-1)/2}=1$ since $P'$ is a group of order $p$, and thus we have the equation from the previous exercise collapsing to $(xy)^p=x^py^p$, and $φ$ is thus a homomorphism.

Assume $φ(x) \not ∈ Z(P)$ for some $x∈P$. This implies $\bar{x}^p≠1$ in $P/Z(P)$, and since $|P/Z(P)|=p^2$, that $\bar{x}$ generates this quotient group, so that it is cyclic, and $P$ is abelian by the familiar process. Therefore, $\text{im}~φ ≤ Z(P)$.

By Cauchy's theorem we have $x$ of order $p$ so that the kernel is not the identity. If the kernel is of order $p$, then the image is of order $p^2$. Since the image being cyclic would imply $|y^p|=p^2$ so $|y|=p^3$ for some $y$, we have the image and thus $P$ contains $p^2-1$ elements of order $p$, and thus the kernel's order is greater than $p$.

As for the oddness of $p$, it was invoked in collapsing the formula in exercise 8. It does indeed map the elements of the only two non-abelian groups of order 8 ($D_8$ and $Q_8$) to their respective centers, but it is not a homomorphism on either, as $φ(s)=1$ and $φ(r)=r^2$, but $φ(rs)=1≠r^2$, and $φ(i)=φ(j)=-1$, but $φ(ij)=-1≠1$. $\square$

The proof to exercise 8 might not be the shortest, though it was not a random long shot. The first induction simplified the problem by replicating the tail of what it ultimately was aiming for and thus appealed to a simpler inductive problem, in a way exhibiting an inductive process itself.

Finite Abelian Groups and Rank (5.2.11)

Dummit and Foote Abstract Algebra, section 5.2, exercise 11:

Let $G$ be a finite abelian group of rank $t$.
(a) Prove that the rank of $G$ is the maximum of the ranks of $G$'s Sylow subgroups.
(b) Prove that the smallest subset of $G$ generating $G$ is of order $t$. [One way of doing this is by combining part (a) with exercise 7.]

Proof: (a) Obtain $G$'s invariant factor decomposition, and let $p$ be a prime dividing $n_s$, the order of the smallest cyclic factor. By 5.1.4, we have the invariant factor decomposition of $P∈\text{Syl}_p(G)$ is of rank $t$.

(b) We know $G$ is generated by the $t$ generators of each of the cyclic factors of $G$, so existence is not an issue. Assume that $G = < S >$, where $| S | = s < t$.

Obtain $P$ of rank $t$ by part (a), and let $φ$ be the $p$-power map. By 5.2.7(a), we have $G/\text{im}~φ ≅ \text{ker}~φ$. Prove that $\text{ker}~φ = \text{ker}~φ|_P$. ($\supseteq$) If $x∈\text{ker}~φ|_P$, then definitionally $x^p=1$ and $x∈ P ≤ G$. ($\subseteq$) Assume $x^p=1$ but $x \not ∈ P$. Since every $p$-subgroup is contained within a Sylow $p$-subgroup, and $P$ is the unique Sylow $p$-subgroup of $G$ by the abelian quality of $G$, we have $< x > ≤ P$, and thus $x∈P$. Therefore $\text{ker}~φ = \text{ker}~φ|_P$. By part (a), we have that $\text{ker}~φ|_P ≅ E_{p^t}$, so that $G/\text{im}~φ ≅ E_{p^t}$, and in particular - as a consequence of the former being generated by the natural projection image of $S$ - the latter is generatable by a subset of fewer than $t$ elements.

For a generating subset $K \subseteq E_{p^t}$ and $| K | = n < t$, we have every element of $E_{p^t}$ is rewritable as $k_1^{\alpha_1}k_2^{\alpha_2}...k_n^{\alpha_n}$. Since every nontrivial element of the elementary abelian group is of order $p$, we have an upper bound of $p^n$ distinct elements writable in this fashion, a contradiction since $p^n < p^t = |E_{p^t}|$. $\square$

The pth-Power Map (5.2.7)

Dummit and Foote Abstract Algebra, section 5.2, exercise 7:

Let $p$ be a prime and let $A = < x_1 > \times < x_2 > \times ... \times < x_n >$ be an abelian $p$-group, where $| x_i | = p^{\alpha_i} > 1$ for all $i$. Define the $p^{th}$-power map: $$φ : A → A~~~\text{by}~~~φ : x \mapsto x^p.$$ (a) Prove that $φ$ is a homomorphism.
(b) Describe the image and the kernel of $φ$ in terms of the given generators.
(c) Prove $\text{ker}~φ ≅ A/\text{im}~φ ≅ E_{p^n}$.

Proof: (a) $φ(ab) = (ab)^{p} = a^pb^p = φ(a)φ(b), ∀a,b∈A$.

(b) We can see that $< x_1^p > \times < x_2^p > \times ... \times < x_n^p > ≤ \text{im}~φ$, so we must prove the reverse inclusion. Let $a∈\text{im}~φ$, so rewrite $a$ thus: $$a = (a_1^p, a_2^p, ..., a_n^p) = ([x_1^{b_1}]^p, [x_2^{b_2}]^p, ..., [x_n^{b_n}]^p)$$ $$= ([x_1^p]^{b_1}, [x_2^p]^{b_2}, ..., [x_n^p]^{b_n})∈< x_1^p > \times < x_2^p > \times ... \times < x_n^p >$$Therefore, $< x_1^p > \times < x_2^p > \times ... \times < x_n^p > = \text{im}~φ$.

Note that $< (x_1)^{p^{\alpha_1-1}} > \times < (x_2)^{p^{\alpha_2-1}} > \times ... \times < (x_n)^{p^{\alpha_n-1}} > ≤ \text{ker}~φ$. Prove the reverse inclusion:$$\text{if}~c∈\text{ker}~φ, \text{then}~c=(c_1, c_2, ..., c_n)=(x_1^{m_1}, x_2^{m_2}, ..., x_n^{m_n})$$Where $(x_i^{m_i})^p=x_i^{m_ip}=1$ for all $i$, so that $p^{\alpha_i} \mid m_ip$, so $p^{\alpha_i-1} \mid m_i$, so $x_i^{m_i}=((x_i)^{p^{\alpha_i-1}})^{v_i}$ for some integer $v_i$. Thus, we have $< (x_1)^{p^{\alpha_1-1}} > \times < (x_2)^{p^{\alpha_2-1}} > \times ... \times < (x_n)^{p^{\alpha_n-1}}>~=\text{ker}~φ$.

(c) A small result to start: If $G$ is a finite abelian group such that for all $g∈G$, we have $g^p = 1$, then $G≅E_{p^n}$ for some $n$. Proof: Observe $G$'s elementary divisor decomposition. If any of its cyclic factors are not of order $p$, then there is an element of order neither $p$ nor $1$, a contradiction. Therefore $G≅Z_p^n=E_{p^n}$.

Now, for any $a∈A$, we have $a^p∈\text{im}~φ$, so for any $\bar{a}∈A/\text{im}~φ$, we have $\bar{a}^p=1$, so $A/\text{im}~φ$ is an elementary abelian group. Likewise, we have $k^p=1$ for all $k∈\text{ker}~φ$, so $\text{ker}~φ$ is an elementary abelian group. We have $A/\text{ker}~φ≅\text{im}~φ$, so that $|A/\text{im}~φ| = |\text{ker}~φ|$. All that remains is proving that $\text{ker}~φ$ is of rank $n$. We know that each of the group factors of $\text{ker}~φ$ has an order of $p$, so therefore $\text{ker}~φ$ is of order $p^n$ and thus isomorphic to $E_{p^n}$. $\square$

Properties of Restricted Direct Products (5.1.17)

Dummit and Foote Abstract Algebra, section 5.1, exercise 17:

Let $I$ be any nonempty index set and $G_i$ be a group for each $i∈I$. The restricted direct product or direct sum $R$ of the groups $G_i$ is the set of elements of the direct product that are the identity in all but finitely many components, i.e. are the nonidentity in finitely many coordinates, which is to say, the set of all elements $\prod a_i \in \prod_{i \in I} G_i$ such that $a_i = 1_i$ for all but a finite number of $i \in I$.
(a) Prove that $R ≤ \prod_{i \in I} G_i$.
(b) Prove that $R \trianglelefteq \prod_{i \in I} G_i$.
(c) Let $I = \mathbb{Z}^+$ and let $p_i$ be the $i^{th}$ prime. Show that if $G_i = \mathbb{Z}/p_i\mathbb{Z}$ for all $i \in \mathbb{Z}^+$, then every element of $R$ of $\prod_{i \in I} G_i$ has finite order, but the unabridged group has elements of infinite order as well. In this case, show that $Tor(\prod_{i \in I} G_i) = R$ (cf. 2.1.6).

Proof: (a) We have $\prod a_i\prod b_i = \prod a_i b_i$ has a nonidentity element in the $j^{th}$ coordinate only if either $\prod a_i$ or $\prod b_i$ has a nonidentity element in that coordinate, which limits the potential coordinates of their product for nonidentity elements to occur down to a finite number, and thus the product is within $R$. Furthermore, inverses clearly don't change nonidentity coordinates, the identity element is within $R$, and the associativity axiom is evident.

(b) For any $\prod c_i \in \prod_{i \in I} G_i$ and $\prod a_i \in R$, we have the $j^{th}$ coordinate of $\prod c_i\prod a_i\prod c_i^{-1}$ is equal to $1$ if $a_j = 1$. This implies that there is only a finite number of coordinates for nonidentity coordinates to occur in, and therefore $\prod c_i\prod a_i\prod c_i^{-1} \in R$ and $\prod c_i R \prod c_i^{-1} = R$ so that $R \trianglelefteq \prod_{i \in I} G_i$.

(c) Let $\prod a_i$ be the element such that $a_i = 1$ for all $i$. If $| \prod a_i | = n < \infty$, then observe the $(n+1)^{th}$ coordinate; we should have $a_{n+1} n ≡ n ≡ 0 \mod p_{n+1}$, even though $0 < n < p_{n+1}$. Therefore $| \prod a_i | = \infty $.

For any $\prod a_i \in R$, set $n = \prod_{i \in K} p_i$, where $K$ is the subset of $I$ such that $a_i$ is nonidentity. For any nonidentity coordinate, we have $a_m n ≡ 0 \mod p_m$. Clearly, it also fixes the identity coordinates. Therefore, $| \prod a_i | \mid n$, and in particular the order of every element is finite. This proves $R ≤ Tor(\prod_{i \in I} G_i)$.

Assume $\prod x_i \in Tor(\prod_{i \in I} G_i)$ such that $\prod x_i$ has infinitely many nonidentity coordinates. Observe its order $n$ and the $(n+1)^{th}$ nonidentity coordinate. We have $x_{m} n ≡ 0 \mod p_{m}$ with $x_{m} \not ≡ 0$, and we can see $0 < n < p_{n+1} ≤ p_{m}$. Therefore, $p_{m} \nmid x_{m}$ and $p_{m} \nmid n$, so that $p_{m} \nmid x_{m} n$, i.e. $x_{m} n \not ≡ 0 \mod p_{m}$, a contradiction. Therefore, $Tor(\prod_{i \in I} G_i) ≤ R$ and now $Tor(\prod_{i \in I} G_i) = R$. $\square$

Introduction to Central Products (5.1.12)

Dummit and Foote Abstract Algebra, section 5.1, exercise 12:

Let A and B be groups, and assume Z₁ ≤ Z(A) is isomorphic to Z₂ ≤ Z(B). Let this isomorphism be given by  x_i → y_i for all x_i∈Z₁. A central product of A and B is a quotient (A × B) / Z where Z = { (x_i,y_i^-1) | x_i∈Z₁ } and is denoted by A ∗ B - it is not unique due to the variability in choice of Z₁, Z₂, and the isomorphism φ between them. (Think of A ∗ B as the direct product of A and B "collapsed" by identifying each element of Z₁ with its corresponding element in Z₂.)
(a) Prove that the images of A and B in the quotient group A ∗ B are isomorphic to A and B, respectively, and that these images intersect in a central subgroup isomorphic to Z₁. Find | A ∗ B |.
(b) Let Z₄ ∗ D₈ be the central product that identifies x² and r² and let Z₄ ∗ Q₈ be the central product that identifies x² and -1. Prove that Z₄ ∗ D₈ ≅ Z₄ ∗ Q₈.

Proof: (a) Let ψ be the function mapping A to its image within the central product. It is a homomorphism by:

ψ(ac) = (ac, 1)Z = (a, 1)(c, 1)Z = (a, 1)Z*(c, 1)Z = ψ(a)ψ(c), and it is clearly surjective on its image. Prove that it is injective thus:

ψ(a) = ψ(c)
(a, 1)Z = (c, 1)Z
(c^-1a, 1) ∈ Z
(c^-1a, 1) = (x_i,y_i^-1), for some x_i∈Z₁,
c^-1a = x_i and y_i^-1 = 1, the latter of which implies x_i = 1 due to the assumed isomorphism, so
c^-1a = 1
a = c

The proof for the isomorphism of B with its image in the central product is parallel.

For the second part of (a), we can prove ψ(Z₁) = ψ(A)∩ψ(B). Prove that ψ(Z₁) ≤ ψ(A)∩ψ(B) by following:

ψ(Z₁) ≤ ψ(A) clearly, and
(x_a, y_a^-1)∈Z for any x_a∈Z₁ definitionally, so
(x_a, 1)Z = (1, y_a)Z = (1, y)Z for some y∈B, so ψ(Z₁) ≤ ψ(B).

Now prove ψ(A)∩ψ(B) ≤ ψ(Z₁):

We have (a, 1)Z = (1, b)Z, so
(a, 1)(x_i,y_i^-1) = (1, b)(x_j,y_j^-1), and in particular
ax_i = x_j, so a∈Z₁, which is to say every element of ψ(A) that is also in ψ(B) is within ψ(Z₁), so ψ(A)∩ψ(B) ≤ ψ(Z₁). We thus have Z₁ ≅ ψ(Z₁) = ψ(A)∩ψ(B).

Furthermore, we have | Z | = | Z₁ | simply by observing the obvious surjective maps from Z to Z₁ and from Z₁ to Z. Thus, | A ∗ B | = | (A × B) / Z | = | A |*| B | / | Z | = | A |*| B | / | Z₁ |.

(b) Note that | (x, 1) | = 4, | (1, r) | = 4, and | (1, s) | = 2 in Z₄ ∗ D₈. We have < (x, 1), (1, r) > is a group of order 4, 8, or 16. It is not 4 as (1, r)∉< (x, 1) > and it is not 16 as 1 ≠ (1, r)² = (x, 1)². Observing that Z₄ ∗ D₈ is of order 16 by the above order formula, we have that < (x, 1), (1, r), (1, s) > is equal either to < (x, 1), (1, r) > or Z₄ ∗ D₈; it must be the latter, as (1, s)∉< (x, 1), (1, r) >.

Note that | (x, 1) | = 4, | (1, i) | = 4, and | (1, j) | = 4 in Z₄ ∗ Q₈. We have < (x, 1), (1, i) > is a group of order 4, 8, or 16. It is not 4 as (1, i)∉< (x, 1) > and it is not 16 as 1 ≠ (1, i)² = (x, 1)². Observing that Z₄ ∗ Q₈ is of order 16 by the above order formula, we have that < (x, 1), (1, i), (1, j) > is equal either to < (x, 1), (1, i) > or Z₄ ∗ Q₈; it must be the latter, as (1, j)∉< (x, 1), (1, i) >.

 We have that the generators of < (x, 1), (1, r), (1, s) > all commute with each other, and (1, r)² = (x, 1)² so that Z₄ ∗ D₈ = { (x^a, n) | 0 ≤ a ≤ 3, n∈{1, r, s, rs} = D₄} (it helps to draw the element chart out), and similarly, the generators of < (x, 1), (1, i), (1, j) > all commute with each other, and (1, i)² = (x, 1)² so Z₄ ∗ Q₈ = { ((x^a, m) | 0 ≤ a ≤ 3, n∈{1, i, j, k} = Q₈ / < -1 > ≅ D₄ }. For each element of Z₄ ∗ Q₈ obtain the isomorphism to Z₄ ∗ D₈ by applying the isomorphism from Q₈ / < -1 > to D₄ to the second coordinate.

∎

Infinite Extrapolation of the Alternating Group (4.6.6)

Dummit and Foote Abstract Algebra, section 4.6, exercise 6:

Let D be the subgroup of the infinite symmetric group consisting of the permutations that move only a finite number of indices (cf. 4.3.17) and let A be the set of all elements σ∈D such that σ acts as an even permutation of the (finite) set of points it moves. Prove that A is an infinite simple group. [Show that every pair of elements of A lie in a finite simple subgroup of A.]

Proof: Take σ₁∈A such that | M(σ₁) | > 4. Letting σ₂ be any other permutation within A, we have < σ₁,σ₂ > ≤ A_{M(< σ1,σ2>)} ≅ A_{| M(< σ1,σ2>) |}, which due to the choice of σ₁ implies that A_{M(< σ1,σ2 >)} is simple. Define G_i to be A_{M(< σ1,σ2,...,σi >)} and we have G₁ ≤ G₂ ≤ ... ≤ A is a chain of ascending simple subgroups whose union is A, which by 4.6.5 establishes the simplicity of A.

∎

Note: This proof is very informally educated, and is likely to be spotted with logical gaps, if it isn't erroneous in itself. Recorded mainly for posterity purposes.

Normal Subgroups of the Symmetric Group (4.6.2)

Dummit and Foote Abstract Algebra, section 4.6, exercise 2:

Find all normal subgroups of S_n for n > 4.

Proof: Assume a proper nontrivial normal subgroup N such that N ≠ A_n (which, due to A_n's maximality, implies A_n is not contained in N). We have N∩A_n is normal in A_n since, for all x∈N∩A_n and a∈A_n, axa^-1∈N (due to N's normality) and axa^-1∈A_n. Therefore, due to A_n's simplicity, we have N∩A_n = 1, so that every nonidentity element of N is of order 2 (since σ²∈A_n) and not in A_n. Assume | N | > 2 such that there are distinct nonidentity x and y (that are their own inverses). Since xy∈A_n, this means xy = 1, so x = y, a contradiction. Therefore |N| = 2, and since N is a union of conjugacy classes and every nonidentity element has some distinct nonidentity conjugate since C_Sn(σ) < S_n due to Z(S_n) = 1, this exhausts all possibilities for N. Thus, the only proper nontrivial normal subgroup in S_n for n > 4 is A_n. Of course, 1 and S_n are also normal in S_n.

∎

Order Limit of Proper Subgroups of the Alternating Group (4.6.1)

Dummit and Foote Abstract Algebra, section 4.6, exercise 1:

Prove that there are no proper subgroups of A_n of index < n for n > 4.

Proof: Reword it to mean no proper subgroup of index ≤ n-1 exists. Let H be such a proper subgroup and let m = | A_n : H | ≤ n-1. Now let A_n act on H by left multiplication. We have the kernel of this action is the largest normal subgroup in H, and since A_n is simple for n > 4, we have the kernel is the identity and A_n maps injectively into S_m. But | A_n | = n!/2 and | S_m | ≤ (n-1)!, and we have that n!/2 ≤ (n-1)! ⇒ n ≤ 2, a contradiction of premises.

∎

Classification of Cyclic-Group-Only Orders (4.5.52-56)

Dummit and Foote Abstract Algebra, section 4.5, exercises 52-56:

52. Suppose G is a finite simple group in which every proper subgroup is abelian. Prove M∩N=1 for distinct maximal subgroups M and N.
53. Prove that if G is a non-abelian group wherein every proper subgroup is abelian then G is not simple. [Let G be a counterexample to this assertion and use exercise 4.3.24 to show that G has more than one conjugacy class of maximal subgroups. Use the method of 4.3.23 to count elements within these conjugates and show that it is greater than | G |.]
54-55. Prove: n is a product of distinct primes p₁p₂...p_r where p_i doesn't divide p_j-1 for all i, j ⇔ All groups of order n are cyclic.
56. Prove that if G is a finite group in which every proper subgroup is abelian, then G is solvable.

Proof: (52) Due to M and N's maximalities, we have N_G(M) = M and N_G(N) = N. Since M and N are abelian, we have that M ≤ N_G(M∩N) and N ≤ N_G(M∩N). Since M ≠ N and M is not a subgroup of N, M has at least one element not in N, so N < < M, N > = G. Therefore we have < M, N > = G ≤ N_G(M∩N), so that M∩N is a normal subgroup. Since G is simple and M∩N ≠ G, we have M∩N = 1.

(53) First note that if M is a maximal subgroup, then gMg^-1 is a maximal subgroup for all g∈G, since if gMg^-1 < H, then M < (g^-1)H(g^-1)^-1. (*)

Consider the set of elements within the conjugates of a maximal subgroup M. According to 4.3.24, we have an element within G that is not within this set. Call it x and proceed by placing the group < x > within a maximal subgroup N (we know that < x > ≠ G, else G is cyclic). We have by the previous exercise that none of the nontrivial elements within the conjugates of M overlap with those of N, since gMg^-1∩fNf^-1 ≠ 1 ⇒ gMg^-1 = fNf^-1 (since these are both maximal by (*) ) ⇒ N = (f^-1g)M(f^-1g)^-1, even though x∈N was defined to not be within any of the conjugates of M. Since the distinct conjugates of M and N all have non-overlapping elements as constituents, and since the number of distinct conjugates of M and N are respectively | G : M | and | G : N | (recall that since G is simple and M and N are maximal, N_G(M) = M and N_G(N) = N), we obtain (| M | - 1)| G : M | + (| N | - 1)| G : N | = 2| G | - | G : M | - | G : N | nonidentity elements. We can't have M = 1 or N = 1, else M ≤ N or N ≤ M, and if | M | > 1 and | N | > 1, then the above equation produces at least | G | + 1 elements including the identity, a contradiction.

(Note that this argument only required the non-cyclicity of G, not necessary the non-abelianness as proposed in the exercise)

(54) (1 ⇒ 2) Let G be a group and | G | = n. We can tell that the order of every subgroup of G is subject to the characteristic prime decomposition above, so that every proper subgroup of G is cyclic by induction and by the preceding exercise (and footnote) G is either cyclic or not simple. If G is not simple, then find a nontrivial proper normal subgroup N, which in these circumstances is cyclic. Proposition 16 implies that | Aut(N) | = φ(| N |) = (p₁ - 1)(p₂ - 1)...(p_k - 1), where the the primes in the decomposition of | G | have been reordered without a loss of generality. Simultaneously, | N_G(N) / C_G(N) | = | G / C_G(N) | = | G |/| C_G(N)| divides (p₁ - 1)(p₂ - 1)...(p_k - 1). If C_G(N) ≠ G, part of this this would imply that at least one prime p_i divides (p₁ - 1)(p₂ - 1)...(p_k - 1). But since p_i doesn't divide p_j-1 for all i, j and p_i is a prime, we have arrived at a contradiction, so that C_G(N) = G, i.e. N ≤ Z(G), and in particular Z(G) > 1. We have that | G / Z(G) | < | G |, so that G / Z(G) is cyclic by induction, and by 3.1.26 G is abelian. By Cauchy's Theorem take x₁, x₂, ..., x_r such that | x_i | = p_i, and by an inductive extrapolation of the process of 4.4.2, < x₁x₂...x_r > = G so that G is cyclic.

(55) (2 ⇒ 1) Disprove the existence of a contradiction by first identifying p and q within the prime decomposition of n such that p | q - 1. Reset n = p^aq^bm where neither p nor q divides m.

By the ending statements regarding groups of semiprime order on page 143, there exists a non-abelian group K of order pq. Let M = Z_p^a-1q^b-1m. Thus K × M is a group of order p^aq^bm = n, and it is non-abelian (take noncommuting x,y∈K and m∈M, so that (x,m)(y,m)≠(y,m)(x,m)) and thus non-cyclic. This violates the premises, and completes the implication.

(56) First assume that G is abelian. Then G / {1} is abelian, thus G is solvable. So assume that G is not abelian, and by exercise 53 there is a nontrivial proper normal subgroup N of G. Any proper subgroup R of G / N is isomorphic to H / N for some proper subgroup H of G, and since H is abelian in G, we have R is abelian in G / N. In this fashion, every proper subgroup of G / N is abelian, and inductively G / N is solvable. N is abelian and thus N / {1} is abelian and N is solvable, thus by the example theorem on page 105 G is solvable.

∎

Generators of Sylow p-Subgroup of Symmetric Groups (4.5.46)

Dummit and Foote Abstract Algebra, section 4.5, exercise 46:

Find generators for the Sylow p-subgroup of S_p², where p is a prime. Show that this is a non-abelian group of order p^p+1.

Proof: First notice that the order of this symmetric group contains the following factors in its order: p, 2p, ..., p², which are all the factors not coprime to p. Thus, the Sylow p-subgroup is of order p^p+1.

Consider the following series of p-cycles: ( 1 ... p ), ( p+1 ... 2p ), ..., ( p(p-1)+1 ... p² ). These all have order p and commute with each other, so we have that the group H they generate is of order p^p. Proceed by imagining a normalizer K of order p possessing a trivial intersection, so that HK = < H, K > is a group and of order p^p+1. As one can see, the group generated by ( 1 p+1 ... p(p-1)+1 )( 2 p+2 ... p(p-1)+2 )...( p 2p ... p² ) fits these prerequisites, as conjugating H with such a permutation will map the individual p-cycles onto each other, it is clearly a group of order p, and every power of it (i.e. every element in the generated group) moves indices in orbits impossible for any element in H so that H∩K = 1. Therefore, < ( 1 ... p ), ( p+1 ... 2p ), ..., ( p(p-1)+1 ... p² ), ( 1 p+1 ... p(p-1)+1 )( 2 p+2 ... p(p-1)+2 )...( p 2p ... p² ) > ∈ Syl_p(S_p²), and this is clearly a non-abelian group as the last generator doesn't centralize any of the others.

∎

One might compare this method to a slightly more abstracted approach outlined at this wonderful blog. In this case, the largest permutation generator he worked with is actually the product of ( 1 ... p ) and my own, so one easily reconciles both proofs' results.

Induction on Groups with Cyclic Sylow 2-Subgroups (4.5.49)

Dummit and Foote Abstract Algebra, section 4.5, exercise 49:

Prove that if | G | = 2ⁿm where m is odd and P∈Syl₂(G) is cyclic then G has a normal subgroup of order m. [Use induction and exercises 11 and 12 in section 2.]

Work by induction, assuming that all groups with a cyclic Sylow 2-subgroup and orders of the form 2^zm for z < n have a normal subgroup of order m. The basis case for a group of order 2m has already been established in the aforementioned exercises.

Let P = < x >, so that x has an order of 2ⁿ. As made apparent in exercise 11, x's image under the regular left representation of G into S_G is an element comprised of m 2ⁿ-cycles. Since m is odd and the parity of an even-ordered cycle is odd, we have that x is an odd permutation. This implies that GA_G = S_G and by the Second Isomorphism Theorem, | S_G : A_G | = 2 = | G : G∩A_G |, so that G∩A_G is a group of order 2^n-1m. Note that x² is an element of order 2^n-1 within G∩A_G, so that Q = < x² > ∈ Syl₂(G∩A_G) is cyclic, and by the inductive assumption, there is a subgroup N of order m in G such that G∩A_G ≤ N_G(N), and the latter is either of order 2^n-1m or 2ⁿm, and in particular is either G∩A_G or G.

We now prove N char G∩A_G so that N is normal in G by a previous exercise. Assume that there is a distinct subgroup M of order m in G∩A_G so that | NM | = | N | | M | / | N∩M |, all of which are odd numbers, so | NM | is of odd order. Since N ≤ NM and N is the largest subgroup of odd order, we have N = NM and M = N, violating the premises. Therefore, N char G∩A_G, so N is normal in G.

∎

Clearly, the same technique used to prove N char G∩A_G can be applied, mutatis mutandis, to prove N char G.