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.