Principle Ideals in Integral Domains (7.4.8)

Dummit and Foote Abstract Algebra, section 7.4, exercise 8:

Let $R$ be an integral domain with $a,b∈R$. Prove $(a)=(b)$ if and only if $a=ub$ for some unit $u$.

Proof: ($\Leftarrow$) We clearly have $a∈(b)=Rb$, and since $u^{-1}a=u^{-1}(ub)=b$, we have $b∈(a)=Ra$. Therefore, $Ra=(a) \subseteq (b)$ and $Rb=(b) \subseteq (a)$ so that $(a)=(b)$. ($⇒$) If either of $a$ or $b$ are zero, then both are zero, and the case clearly holds, so assume $a≠0≠b$. We must have $a=ub$ and $va=b$ for some $u,v∈R$. We can compare the two to obtain $va=vub=b$, so that $vub-b=(vu-1)b=0$, and now $vu-1=0$ implying $vu=uv=1$, i.e. $u$ is a unit as claimed.$~\square$

Nilpotency and the Augmentation Ideal (7.4.3)

Dummit and Foote Abstract Algebra, section 7.4, exercise 3:

(a) Let $G$ be an abelian group of order $p^n$. Prove $Aug(\mathbb{F}_pG)=\mathfrak{N}(\mathbb{F}_pG)$.
(b) Let $G=\{g_1,...,g_n\}$ be a group and assume $R$ is commutative. Prove $r∈Aug(RG)⇒r(g_1+...+g_n)=0$.

Proof: (a) ($\subseteq$) By the preceding exercise, $Aug(\mathbb{F}_pG)$ is generated by $\{g-1~|~g∈G\}$, so for arbitrary $x∈Aug(\mathbb{F}_pG)$ write $x=\sum_{g∈G}a_g(g-1)$ for some $a_g∈\mathbb{F}_pG$ for all $g∈G$. Prove $(g-1)^{p^n}=0$ and is thus nilpotent for any $g∈G$, so that by the idealness of $\mathfrak{N}(\mathbb{F}_pG)$ we have $x$ is nilpotent. By the binomial theorem for commutative rings, since $p^n$ divides $p^n \choose k$$= \dfrac{p^n!}{k!(n-k)!}$ when $0 < k < p^n$, we can observe $(g-1)^{p^n}=g^{p^n}-1=1-1=0$. ($\supseteq$) Since $\mathbb{F}_pG/Aug(\mathbb{F}_pG)≅\mathbb{F}_p$ is a field, we have $Aug(\mathbb{F}_pG)$ is a maximal ideal, so that $\mathfrak{N}(\mathbb{F}_pG)∈\{Aug(\mathbb{F}_pG),\mathbb{F}_pG\}$, and is evidently not the latter as $1∈\mathbb{F}_pG$ is not nilpotent.$~\square$

(b) Once again write $r=r'(g-1)$ for some $r'∈RG$, and recalling that the action of multiplication on a group by one of its elements is a permutation of that group's elements, we observe $$r(g_1+...+g_n)=r'(g-1)(g_1+...+g_n)=r'(g(g_1+...+g_n)-(g_1+...+g_n))=$$ $$r'((g_1+...+g_n)-(g_1+...+g_n))=r'0=0~\square$$

Nonisomorphicity of Z[x] and Q[x] (7.3.2)

Dummit and Foote Abstract Algebra, section 7.3, exercise 2:

Prove $\mathbb{Z}[x] \not ≅ \mathbb{Q}[x]$.

Proof: Lemma 1: If $φ:R→S$ is a surjective ring homomorphism and $R$ is generated by a subset $X$ (i.e. every element of $R$ is writable as a sum of products with subtraction [or a series of operations] of elements from $X$), then $S$ is generated by $φ(X)$. Proof: Let $s$ be an arbitrary element of $S$. For some $r∈R$, we have $s=φ(r)$ and since $r$ is a series of operations of elements from $X$, due to the homomorphism's nature we have $s$ is a series of operations of elements from $φ(X)$.$~\square$

We can see that $\mathbb{Z}[x]$ is generated by $\{x,1\}$, since the latter generates $\mathbb{Z}$ (all potential coefficients) and the former generates $x^n$ for all $n∈\mathbb{Z}^+$. This implies $\mathbb{Q}[x]$ is generated by two elements, and in particular $\mathbb{Q}$ is generated by two elements (since we have a surjective homomorphism $φ:\mathbb{Q}[x] → \mathbb{Q}$ by $p(x) \mapsto p(0)$). Letting $\dfrac{a}{b}$ and $\dfrac{c}{d}$ denote these generators, we have $ad \cdot \dfrac{1}{bd} = \dfrac{a}{b}$ and $bc \cdot \dfrac{1}{bd}=\dfrac{c}{d}$, so that in fact $\mathbb{Q}$ is generated by one element; further refer to it as $\dfrac{1}{z}$. Letting $p \not \mid z$ be prime, we must have $\dfrac{1}{p}$ is generated by a series of operations of $\dfrac{1}{z}$ despite the fact that every such operation can be condensed into some fraction $\dfrac{y}{z^n}$, which cannot simplify to $\dfrac{1}{p}$.$~\square$

A Ring of Homomorphisms (7.1.30)

Dummit and Foote Abstract Algebra, section 7.1, exercise 30:

Let $A=\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} ...$ be the direct product of copies of $\mathbb{Z}$ indexed by positive integers and let $R$ be the ring of homomorphisms from $A$ to itself. Let $φ$ be the element of $R$ defined by $φ(a_1,a_2,a_3,...)=(a_2,a_3,...)$. Let $ψ$ be the element of $R$ defined by $ψ(a_1,a_2,a_3,...)=(0,a_1,a_2,a_3,...)$.
(a) Prove $φψ=1≠ψφ$.
(b) Exhibit infinitely many right inverses for $φ$.
(c) Find a nonzero $\pi∈R$ such that $φ\pi = 0 ≠ \piφ$.
(d) Prove $\nexists \lambda∈R~(\lambdaφ=0)$.

Proof: (a) $$(φψ)(a_1,a_2,a_3,...)=φ(ψ(a_1,a_2,a_3,...))=φ(0,a_1,a_2,...)=(a_1,a_2,a_3,...)$$ (b) Let $\pi$ be the homomorphism defined in part (c). We claim $\pi^n+ψ$ is a right inverse for any $n∈\mathbb{Z}^+$. We have $φ(\pi^n+ψ)=φ\pi\pi^{n-1}+φψ=φψ=1$, so now it suffices to prove $\pi^n+ψ≠\pi^m+ψ$ for any $m≠n$: $$\pi^n+ψ=\pi^m+ψ⇒\pi^n=\pi^m⇒$$ $$\pi^n(1,0,...)=\pi^m(1,0,...)⇒(2^n,0,...)=(2^m,0,...)⇒n=m$$ (c) Let $\pi$ be defined by $\pi(a_1,a_2,a_3,...)=(2a_1,0,0,...)$. Prove it is a homomorphism: $\pi((a_1,a_2,...)+(b_1,b_2,...))=\pi((a_1+b_1,a_2+b_2,...)=(2(a_1+b_1),0,...)=$$(2a_1,0,...)+(2b_1,0,...)=\pi(a_1,a_2,...)+\pi(b_1,b_2,...)$. We have $(φ\pi)(a_1,a_2,...)=φ(\pi(a_1,a_2,))=φ(2a_1,0,...)=(0,0,...)$, so that $φ\pi=0$. Now, $(\piφ)(0,1,0,0,...)=(2,0,0,...)$, so that $\piφ≠0$.
(d) By exercise 28(b), since $φ$ has $ψ$ a right inverse, $φ$ is not a right zero divisor.$~\square$

Order of Conductor f (7.1.23)

Dummit and Foote Abstract Algebra, section 7.1, exercise 23:

Define $\mathcal{O}_f=\mathbb{Z}[f\omega]=\{a+b\omega~|~a,b∈\mathbb{Z}\}$. Letting $[R]$ denote the abelian group in the ring $R$, prove: $$R = \mathcal{O}_f ⇔ R~\text{subring of}~\mathcal{O} \land 1∈R \land |[\mathcal{O}] : [R]|=f$$ Proof: Lemma 1: For $H≤G$ groups and $φ$ an isomorphism, we have $|G:H|=|φ(G):φ(H)|$. Proof: Let $n_1,...$ be a complete set of coset-distinct representatives for $H$ in $G$. We claim that their images are a complete set of coset-distinct representatives of $φ(H)$ in $φ(G)$. For any $g∈G$, we have $g=n_ih$ for some index $i$ and $h∈H$. Therefore, we have $φ(g)=φ(n_ih)=φ(n_i)φ(h)$. Assuming $φ(n_i)$ and $φ(n_j)$ are in the same coset of $φ(H)$, we have $φ(n_i)=φ(n_j)φ(h)$$=φ(n_jh)$, so that $n_i=n_jh$ and $n_i$ and $n_j$ are in the same coset of $H$, a contradiction. Thus, there are just as many coset representatives of $φ(H)$ in $φ(G)$ as $H$ in $G$.$~\square$

($⇒$) (1) All that needs to be done is to prove closure under subtraction and multiplication. We have $(a+bf\omega)-(c+df\omega)=(a-c)+(b-d)f\omega$. As well, $\omega^2=D$ or $\dfrac{D-1}{4}+\dfrac{1+\sqrt{D}}{2}$ depending on $D$ modulo 4, so $$(a+bf\omega)(c+df\omega)=(ac)+(bc+ad)f\omega+bdf^2\omega^2$$ is evidently within the ring as well. (2) Clearly $1∈\mathcal{O}_f$; set $a=1$ and $b=0$. (3) Construct a mapping $φ:[\mathcal{O}]→\mathbb{Z} \times \mathbb{Z}$ by $a+b\omega \mapsto (a,b)$. This is well defined by unique $a,b$ form of the argument, clearly surjective and injective, and $$φ((a+b\omega)+(c+d\omega))=φ((a+c)+(b+d)\omega)=(a+c,b+d)$$ $$=(a,b)+(c,d)=φ(a+b\omega)+φ(c+d\omega)$$ so that $φ$ is homomorphic and now an isomorphism. So $[\mathcal{O}]≅\mathbb{Z} \times \mathbb{Z}$ and by definition we can tell $[\mathcal{O}_f] ≅ φ([\mathcal{O}_f])=\mathbb{Z} \times f\mathbb{Z}$. By 5.1.14, we have $(\mathbb{Z} \times \mathbb{Z})/(\mathbb{Z} \times f\mathbb{Z})≅$$(\mathbb{Z}/\mathbb{Z})\times(\mathbb{Z}/f\mathbb{Z})≅\mathbb{Z}/f\mathbb{Z}$ is a group of order $f$, so that $f=|\mathbb{Z} \times \mathbb{Z} : \mathbb{Z} \times f\mathbb{Z}|=|φ([\mathcal{O}]):φ([\mathcal{O}_f])|=|[\mathcal{O}]:[\mathcal{O}_f]|$.

Lemma 2: Any nontrivial subgroup of $\mathbb{Z}$ is of the form $n\mathbb{Z}$ for some $n∈\mathbb{Z}^+$. Proof: Let distinct $a,b$ be chosen with $n=|a-b|$ minimal; we thus have $n\mathbb{Z}$ is contained within the subgroup. Assume there is an element $x$ not within $n\mathbb{Z}$: We have $x=qn+r$ with $0 < r < n$, and since $qn∈n\mathbb{Z}$, we have $x-qn < n=|a-b|$, a contradiction. So $n\mathbb{Z}$ is precisely the subgroup in question.$~\square$

($\Leftarrow$) $R$'s identity must be $1$, as $n(a+bf\omega)≠a+bf\omega$ for any $n≠1$ and $a≠0≠b$. Therefore, we have $1\mathbb{Z} \times 0=\mathbb{Z} \times 0≤φ([R])$. For any $(a,b)∈φ([R])$, we also have $(a,0)∈φ([R])$, so that $(0,b)∈φ([R])$, implying the second factor of $φ([R])$ is a subgroup of $\mathbb{Z}$ and therefore of the form $n\mathbb{Z}$. Since $| \mathbb{Z} \times \mathbb{Z}:\mathbb{Z} \times n\mathbb{Z}|=n$, we have $n=f$ and now $φ([R])=\mathbb{Z} \times f\mathbb{Z}=φ([\mathcal{O}_f])$ and now $[R]=[\mathcal{O}_f]$. Since the specifications of multiplication are predefined in $\mathbb{C}$, we have $R=\mathcal{O}_f$.$~\square$

Commutativity of Boolean Rings (7.1.15-16)

Dummit and Foote Abstract Algebra, section 7.1, exercises 15-16:

15. A ring $R$ is a Boolean ring if $a^2=a$ for all $a∈R$. Prove that every Boolean ring is commutative.
16. Prove that the only Boolean integral domain is $\mathbb{Z}/2\mathbb{Z}$.

Proof: (15) Take any $a,b∈R$. Note that we necessarily have $(-x)^2=-x$ for any $x∈R$ as well as $(-x)^2=(-x)(-x)=x^2=x$, so that $-x=x$. Now, notice that $(a+b)^2=a+b$ by the Boolean property. By distributing, we have $(a+b)^2=a^2+ab+ba+b^2=a+ab+ba+b$. Comparing the two, we have $ab+ba=0$, so that $ab=-ba=ba$.$~\square$

(16) For any $a$ in a Boolean integral domain, we have $a(a-1)=a^2-a=a-a=0$, so that either $a=0$ or $a-1=0$ implying $a=1$.$~\square$

Free Groups and Nilpotency/Solvability (6.3.12)

Dummit and Foote Abstract Algebra, section 6.3, exercise 12:

Let $S$ be a set and let $c$ be a positive integer. Formulate the notion of a free nilpotent group on $S$ of nilpotence class $c$ and prove it has the appropriate universal property with respect to nilpotent groups of class $≤c$.

Proof: Let $N$ be any group of nilpotence class $≤c$ and let $ψ: S → N$ be a set map with $G = \langle~\text{img}~ψ~\rangle$. We have a unique homomorphism $φ : F(S) → N$ (fixing $S$) so that $F(S)/\text{ker}~φ ≅ G$. We prove that there is a unique homomorphism $\Phi : F(S)/F(S)^c → N$ such that $\Phi\mid_{\pi(S)} = ψ$ (where $\pi$ is the natural homomorphism from $F(S)$ to $F(S)/F(S)^c$). Assume $F(S)^c \not ≤ \text{ker}~φ$; we then have a contradiction: $$G^c≅(F(S)/\text{ker}~φ)^c=F(S)^c/\text{ker}~φ≠1$$ Therefore there is the desired homomorphism afforded by: $$(F(S)/F(S)^c)/(\text{ker}~φ/F(S)^c)≅F(S)/\text{ker}~φ≅G$$ Assume $\Phi_1 ≠ \Phi_2$ are two homomorphisms from $F(S)/F(S)^c$ to $N$ fixing $\pi(S)$; then $\Phi_1 \circ \pi ≠ \Phi_2 \circ \pi$ are two homomorphisms from $F(S)$ to $G$ fixing $S$, a contradiction. We thus have $\text{ker}~φ$ factors through $F(S)^c$, and the following diagram commutes:$\square$
Note that this theorem can be paralleled to produce a similar result regarding a free solvable group on a set $S$.

Presentation of the Quaternion Group (6.3.7)

Dummit and Foote Abstract Algebra, section 6.3, exercise 7:

Prove the following is a valid presentation: $$Q_8 = \langle~a,b~\mid~a^2=b^2,~a^{-1}ba=b^{-1}~\rangle$$ Proof: By the first, we have $a^2b^{-2}=1$. However, by the second, we also have $a^2b^{-2}=a^2(a^{-1}ba)(a^{-1}ba)=ab^2a$, so that $ab^2a=1$, then $a^{-2}=b^2=a^2$ so $a^4=1$. Now, $a^2=b^2⇒a^4=1=b^4$.

Prove by induction that $b^ka=ab^{-k}$. The case holds for $k=1$ by the presentation, and $b^{k+1}a=bb^ka=bab^{-k}=ab^{-1}b^{-k}=ab^{-(k+1)}$. By extension, $ab^k=b^{-k}a$. We can see by repeated multiplication the left by $a$, that generally $a^ib^k=b^{\pm k}a^i$.

Attempt to observe any element with a reduced width $n$ greater than 2, necessarily in either the form $...a^{e_{n-2}}b^{e_{n-1}}a^{e_n}=...a^{e_{n-2}+e_n}b^{\pm e_{n-1}}$ or $...b^{e_{n-2}}a^{e_{n-1}}b^{e_n}=...b^{e_{n-2} \pm e_n}a^{e_{n-1}}$, a contradiction in any case. So every element can be reduced to the form $a^xb^y$ or $b^xa^y$ (with $0 ≤ x,y ≤ 3$), the latter of which can be ignored as $b^xa^y=a^yb^{\pm x}$. Using the relations we have been given and have established, we can easily establish equivalency classes among these sixteen candidates to provide a maximum of eight distinct elements (e.g. $ab^3=aa^2b=a^3b$). Since $i,j∈Q_8$ fulfill the relations presented above for $a,b$ and $\langle~i,j~\rangle = Q_8$, we have the presentation validated.$~\square$

Linear Systems of Two Variables and Equivalency (1.2.6)

Hoffman and Kunze Linear Algebra, section 1.2, exercise 6:

Let $F_1$ and $F_2$ be homogeneous systems of linear equations in two unknowns. Prove that if their solution sets are identical, then they are equivalent.

Proof: Lemma 1: Admit two linear systems with one equation each, $A : ax_1+bx_2=0$ and $B : cx_1+dx_2=0$. We have $A$ equivalent to $B$ if and only if $A$ and $B$ have the same solution set if and only if $\dfrac{a}{c}=\dfrac{b}{d}$. Proof: (1 ⇒ 2) Theorem 1 ensures this. (2 ⇒ 3) Any solution $(n_1,m_1)$ of $A$ must satisfy $(n_1,-\dfrac{an_1}{b})$, and this is seen to be a valid solution so these tuples form the solution set of $A$. Similarly, the solution set of $B$ is of the form $(n_2,-\dfrac{cn_2}{d})$. Due to these solution sets' equivalence, we have $(1,-\dfrac{c}{d})$ is a solution of $A$, so that $a(1)+b(-\dfrac{c}{d})=0$, entailing (3). (3 ⇒ 1) Let $c_1 = \dfrac{c}{a} = \dfrac{d}{b}$. Evidently $c_1(ax_1+bx_2)=cx_1+dx_2=c_1(0)=0$.$~\square$

Lemma 2: Let $ax_1+bx_2=0$ and $cx_1+dx_2=0$ be two equations that aren't scalar products of each other. We have their shared solution set is $(0,0)$. Proof: The solution sets of both are $(n_1,-\dfrac{an_1}{b})$ and $(n_2,-\dfrac{cn_2}{d})$ respectively. Assume they share a nontrivial solution. Then $n_1=n_2≠0$ and $-\dfrac{an_1}{b}=-\dfrac{cn_2}{d}=-\dfrac{cn_1}{d}$, entailing $\dfrac{a}{c}=\dfrac{b}{d}$ and these two equations are scalar products by the first lemma, a contradiction.$~\square$

Let $F_1$ contain $n$ equations and $F_2$ contain $m$ equations. Assume $\dfrac{A_{j_1 1}}{A_{k_1 1}}=\dfrac{A_{j_1 2}}{A_{k_1 2}}$ and $\dfrac{B_{j_2 1}}{B_{k_2 1}}=\dfrac{B_{j_2 2}}{B_{k_2 2}}$ for all $1 ≤ j_1,k_1 ≤ n$ and $1 ≤ j_2,k_2 ≤ m$, i.e. all the equations of $F_1$ are scalar products of each other, and the same pertaining to $F_2$, as by the first lemma. Now, $(n,-\dfrac{nA_{1 1}}{A_{1 2}})$ is the solution set of $F_1$. Let $B_{v 1}x_1 + B_{v 2}x_2 = 0$ be an arbitrary equation of $F_2$. We have $(1,-\dfrac{A_{1 1}}{A_{1 2}})$ is a solution of this equation, so that $B_{v 1}+B_{v 2}(\dfrac{A_{1 1}}{A_{1 2}})=0$ and now $\dfrac{A_{1 1}}{B_{v 1}}=\dfrac{A_{1 2}}{B_{v 2}}$ so that by the first lemma this arbitrary equation of $F_2$ is a scalar product of the equation from $F_1$, and now $F_2$ is a linear combination of $F_1$. Reversing the argument, we obtain $F_1$ is a linear combination of $F_2$, and now $F_1$ and $F_2$ are equivalent.

So assume that there exist two equations in one of the systems that are not scalar products of each other, and designate their system $F_1$. These equations are $ax_1+bx_2=0$ and $cx_1+dx_2=0$, and evidently at least one of $a$ or $c$ must be nontrivial for their assumption to hold; let it be $c$. Let $nx_1+mx_2=0$ be an arbitrary equation of $F_2$. This equation is a linear combination of the two former by scalars $c_1$ and $c_2$ if (and only if) $c_1a+c_2c=n$ and $c_1b+c_2d=m$; solving these two nonhomogenous equations, we obtain $$c_1=\dfrac{m-\dfrac{dn}{c}}{b-\dfrac{da}{c}}~\text{and}~c_2=\dfrac{n-c_1a}{c}$$ are such solution scalars. These avoid division by zero as $c≠0$ and $b-\dfrac{da}{c}=0 ⇒ \dfrac{a}{c}=\dfrac{b}{d}$, a contradiction by the first lemma and this paragraph's assumption. Thus $F_2$ is a linear combination of $F_1$.

$F_1$ having two mutually non-scalar-product equations implies the same for $F_2$, since by the second lemma the solution set of $F_1$ is $(0,0)$, the same applies for $F_2$, yet the first lemma would entail a nontrivial solution set for $F_2$ if its equations were all scalar products of each other in two unknowns. Thus the argument above can be reversed to obtain $F_1$ is a linear combination of $F_2$ and now $F_1$ is equivalent to $F_2$.$~\square$

Free Groups and Rank (6.3.1)

Dummit and Foote Abstract Algebra, section 6.3, exercise 1:

Let $\langle~a_1,...,a_n~\rangle=F_1$ and $\langle~b_1,...,b_m~\rangle=F_2$ be free groups. Prove that $F_1≅F_2$ if and only if $n=m$. Consider the case when the two's ranks are infinite, as well.

Proof: ($⇒$) Assume $n≠m$, and generally $n$ is of smaller cardinality than $m$. Let $φ$ be the isomorphism from $F_1$ to $F_2$. For an arbitrary element $x∈F_2$, we have $x=φ(a_1^{k_1}...a_n^{k_r})=φ(a_1)^{k_1}...φ(a_r)^{k_r}$, so that $F_2$ is generated by a number of elements of smaller than or equal cardinality to those of $F_1$, a contradiction. ($\Leftarrow$) Let there be a bijection between the generators of $F_1$ and the generators of $F_2$, operating on the index by $\pi$. Define a mapping $φ : F_1 → F_2$ by $φ(a_i)=b_{\pi(i)}$ and homomorphically extend it. For any two generators in $F_1$, we have: $$φ(a_i)^x=φ(a_j)^y⇒b_{\pi(i)}^x=b_{\pi(j)}^y⇒\pi(i)=\pi(j)⇒i=j⇒a_i=a_j$$ Now observe: $$φ(a_1^{k_1}...a_r^{k_r})=φ(a_1^{j_1}...a_s^{j_s})⇒b_{\pi(1)}^{k_1}...b_{\pi(r)}^{k_r}=b_{\pi(1)}^{j_1}...b_{\pi(s)}^{j_r}⇒∀i[b_{\pi(i)}^{k_i}=b_{\pi(i)}^{j_i}]$$ $$⇒∀i[φ(a_i)^{k_i}=φ(a_i)^{j_i}]⇒∀i[a_i^{k_i}=a_i^{j_i}]⇒a{\pi(1)}^{k_1}...a_{\pi(r)}^{k_r}=a_1^{j_1}...a_s^{j_s}$$ So that $φ$ is an injective homomorphism. Similarly, we can construct an injective homomorphism from $F_2$ into $F_1$, therefore providing an isomorphism. This argument did not assume the finiteness of rank, merely the equivalency of cardinality, so is therefore applicable to infinite groups as well.$~\square$

Warning: Invalid proof and erroneous extension of the Bernstein-Schroeder theorem.

The Millionaire (6.2.28)

Dummit and Foote Abstract Algebra, section 6.2, exercises 28:

Let $G$ be simple and of order $3^3*7*13*409=1,004,913$. Calculate the number of Sylow $p$-subgroups for each prime dividing $|G|$.

Proof: Preliminary Sylow analysis shows: $$n_3∈\{7,13,91,409,2863,5317,37219\}~~~~~n_7∈\{351,47853\}$$ $$~~~~~n_{13}∈\{27,25767\}~~~~~n_{409}∈\{819\}$$ Removing the numbers that violate the index restrictions, we have all of them solved except for $n_3$. Assume $n_3=409$, and we have $|N_G(P_3)|=3^3*7*13$, so for some $P_7$ we have $P_3P_7$ is a subgroup of $N_G(P_3)$, so that naturally $P_7 \trianglelefteq P_3P_7$, meaning $|P_3P_7|=189 \mid |N_G(P_7)|=21$, a contradiction. Assuming $n_3=5317$, we end up with the same contradiction as then $|N_G(P_3)|=3^3*7$.

A brief digression: By counting the elements of order 7, 13, and 409, we obtain $930,474$ elements. If there is an element $x$ of order 21, then $P_7 \trianglelefteq \langle~x~\rangle$ so that $\langle~x~\rangle = N_G(P_7)$ by order, so that by 6.2.16's first lemma there are $12*47853=574,236$ elements of order 21, overloading $G$.

Now assume $n_3=37219$, and since $37129 \not ≡ 1 \mod{3^2}$, we have an order-$3^2$ intersection of two Sylow 3-subgroups $P_3∩Q_3$, whose normalizer is of order $3^3*7$,$3^3*13$, or $3^3*7*13$. If $3^3*7$, then by the above the only order the elements in this normalizer can take are 1, 7, and powers of 3. Sylow shows $n_3=7$ (by 6.2.13) and $n_7=1$, so that there are $3^3*7-6=183$ elements in the Sylow 3-subgroups. Without taking intersections into account, there is a maximum of $7*26+1=183$ elements possible, so that there is no intersection between Sylow 3-subgroups taking place, despite $P_3,Q_3 ≤ N_G(P_3∩Q_3)$ and $P_3∩Q_3 ≠ 1$, or 6.2.13 in general. If $3^3*13$ or $3^3*7*13$, the order of the normalizer forces only one Sylow 3-subgroup, despite once again $P_3,Q_3 ≤ N_G(P_3∩Q_3)$ being distinct Sylow 3-subgroups. Therefore, $n_3=2863$ by default.$~\square$

Groups of Order 168 (6.2.26)

Dummit and Foote Abstract Algebra, section 6.2, exercise 26:

Evaluate the validity of the following statement: $$|G|=168 \land~n_7(G)>1 ⇒ G~\text{is simple}$$ Proof: The statement is false. Let $Z_2^3 \rtimes Z_7$ denote the semidirect product afforded by a homomorphism of the generator of $Z_7$ into a generator of a cyclic Sylow 7-subgroup of $Aut(Z_2^3)≅GL_3(Z_2)$ of order $168=2^3*3*7$. This implies $Z_7 \not \trianglelefteq Z_2^3 \rtimes Z_7$, so that $n_7(Z_7 \rtimes Z_2^3)>1$. Now, observe the group $G=(Z_2^3 \rtimes Z_7) \times Z_3$ of order 168. We have the former group as a subgroup of this one, so that $n_7(G)>1$. Furthermore, if $((a,b),c)∈P_3∈Syl_3(G)$, then $((a,b),c)^3=((a,b)^3,c^3)=((1,1),1)$, and since $(a,b)$ is an element of the subgroup of order $56$, we must have $(a,b)=(1,1)$ so that there are only $|Z_3|=3$ distinct elements of $G$ satisfying $x^3=1$, therefore $n_3(G)=1$, $P_3 \trianglelefteq G$ and $G$ is not simple.$~\square$

Restrictions of Groups of Small Order (6.2.18-20)

Dummit and Foote Abstract Algebra, section 6.2, exercises 18-20:
18. Prove $|G|=36⇒n_2 = 1 \lor n_3 = 1$.
19. Prove $|G|=12~\land \nexists H(H ≤ G \land |H| = 6)⇒G≅A_4$.
20. Prove $|G|=24~\land \nexists g(g∈G \land |g| = 6)⇒G≅S_4$.

Proof: (18) Sylow analysis reveals: $$n_2∈\{3,9\}~~~~~n_3=4$$ Let $G$ act by left multiplication on the four cosets of $N_G(P_3)=P_3$; the kernel $N$ of this action is the largest normal subgroup of $G$ contained in $P_3$. Since $P_3 \not \trianglelefteq G$ and $N=1$ allows $G$ isomorphic passage into $S_4$ of order 24, we must have $|N|=3$. We have $|G/N|=12$, so there is either a normal Sylow 2-subgroup or a normal Sylow 3-subgroup of $G/N$. Assuming the latter, we have $K/N \trianglelefteq G/N$ is of order 3, so that its preimage $K$ is of order 9 and normal in $G$, a contradiction. Therefore let $K/N \trianglelefteq G/N$ be of order 4 so that $K \trianglelefteq G$ is of order 12. It contains some Sylow 2-subgroup of $G$, and by its normality it contains all of them, so $n_2=3$. Let $K$ act by left multiplication on the three cosets of $P_3$ it contains; once again, we must have a normal subgroup $T$ of order 2 in $K$. This can only be the 2-core of $K$, which is characteristic in $K$, and now normal in $G$. We have $TN$ is a normal subgroup of order 6, so that $P_3(TN)$ is a subgroup; since $P_3 \not ≤ TN$ and $P_3∩TN=1$ implies $P_3(TN)$ is a subgroup of order 54, we must have $|P_3∩TN|=3$ and $P_3(TN)$ is a subgroup of order 18. This subgroup is normal by its index, contains a Sylow 3-subgroup and thus contains all of them, and yet Sylow purports $n_3=1$.$~\square$

(19) Since $G \not ≅ A_4$ by assumption, we have $n_3=1$. Take $x$ of order 2 and we have $\langle~x~\rangle P_3$ is a subgroup of order 6.$~\square$

(20) We have: $$n_2∈\{1,3\}~~~~~n_3∈\{1,4\}$$ If $n_3=4$, then letting $G$ act on the cosets of $P_3$ allows isomorphic passage into $S_4$ so that by order $G≅S_4$. In addition, assume $n_2=1$; then $G=P_2P_3 ≅ P_2 \times P_3$. If $Z(P_2)=P_2$, then $Z(G) ≅ Z(P_2) \times Z(P_3) = P_2 \times P_3$, so that $G$ is abelian and the elements $x$ and $y$ of orders 2 and 3 respectively combine for a product $xy$ of order 6. If $|Z(P_2)|=2$ (the only other option for $Z(P_2)$), then $Z(P_2)~\text{char}~P_2 \trianglelefteq G$ so that $Z(P_2) \trianglelefteq G$ and now $Z(P_2)P_3 ≅ Z_6$. Ultimately, $n_2=3$. Since $3 \not ≡ 1 \mod{2^2}$ we have $P_2∩Q_2$ of order 4; by exercise 13, we must have $N_G(P_2∩Q_2)=G$ so that $P_2∩Q_2 \trianglelefteq G$, and by 4.5.37 $P_2∩Q_2$ is contained in every Sylow 2-subgroup, so that the sum of the distinct elements of order 2, 4, and 8 can be explicitly calculated; there is the intersection point of order 4 minus the identity, and the three Sylow 3-subgroups that each provide 4 unique elements (lest the intersection between two distinct Sylow 2-subgroups be greater than 4, an impossibility), yielding 15 such elements. Comparing this with the following chart of possible orders of elements that can be determined thus far:we invariably end up with only 18 elements of $G$, a contradiction.$~\square$

Search for Simplicity (6.2.16)

Dummit and Foote Abstract Algebra, section 6.2, exercise 16:

Prove there are no simple groups of odd composite order < 10,000.

Proof: I will record some difficult cases:

1575: Sylow analysis gives us $n_5=21$. Since $21≠1 \mod{5^2}$, we have $P_5∩Q_5$ of order 5 and due to index crunching we obtain $|N_G(P_5∩Q_5)|=3*5^2$. But now $n_5(N_G(P_5∩Q_5))=1$, even though $P_5≠Q_5$ and $P_5,Q_5∈Syl_5(N_G(P_5∩Q_5))$.

3465: Lemma 1: Let $G$ be a group, let $P∈Syl_p(G)$ and assume $N_G(P)$ is cyclic. Then there are precisely $n_p \cdot φ(|N_G(P)|)$ elements of order $|N_G(P)|$ in $G$. Proof: Since $P \trianglelefteq N_G(P)$, we have distinct normalizers of Sylow p-subgroups for distinct Sylow p-subgroups, so that there are $n_p$ distinct normalizers. As well, if $|x| = |N_G(P)|$, then for some $Q∈Syl_p(G)$ we have $Q \trianglelefteq \langle~x~\rangle$ so that $\langle~x~\rangle=N_G(Q)$, and now every element of such an order is contained in a normalizer. Since these normalizers are conjugate to one another and are thus cyclic as well, we have each normalizer contains $φ(|N_G(P)|)$ elements of the specified order, and since no two distinct normalizers share elements of this order, we have the lemma proven.$~\square$

Lemma 2: A group $G$ of order $231=3*7*11$ has an element of order 33. Proof: We have $P_{11} \trianglelefteq G$, so that $P_3P_{11} ≅ Z_{33}$ is a subgroup of $G$.$~\square$

By Sylow analysis on 3465, we obtain: $$n_3∈\{7,55,385\}~~~~~n_5∈\{11,21,231\}~~~~~n_7∈\{15,99\}~~~~~n_{11}=45$$ This reveals that $N_G(P_{11}) ≅ Z_{77}$ (in particular, this implies $G$ has no elements of order 33). By the first lemma, there are $45*60=2700$ elements of order 77 in $G$. By $n_{11}=45$, we obtain 450 elements of order 11. Now, if $n_7=99$, this would produce 594 elements of order 7, which overloads the order of $G$. Therefore $n_7=15$, and now $|N_G(P_7)|=231$ and contains an element of order 33 by the second lemma, a terminating contradiction.

9765: By Sylow analysis and index crunching, we obtain: $$n_3∈\{31,217\}~~~~~n_5∈\{31,651\}~~~~~n_7=155~~~~~n_{31}=63$$ We have $|N_G(P_7)|=3^2*7$. Now, if $n_5=31$, we have $|N_G(P_5)|=3^2*5*7$, so that $P_7 \trianglelefteq P_5P_7$ and then $35 \mid |N_G(P_7)|=63$, a contradiction. So $n_5=651$ and $N_G(P_5)≅Z_{15}$, so that by lemma 1 we have 5208 elements of order 15. With Sylow, we obtain at least $5208+651*4+155*6+63*30=10,632$ elements in $G$, an impossibility.$~\square$

Normalizers of Sylow Intersections (6.2.13)

Dummit and Foote Abstract Algebra, section 6.2, exercise 13:

Let $G$ be a group with more than one Sylow $p$-subgroup. Choose $P,Q∈Syl_p(G)$ such that $|P \cap Q|$ is maximal. Prove that $N_G(P∩Q)$ has more than one Sylow $p$-subgroup, any two distinct Sylow $p$-subgroups intersect in $P∩Q$, and $p \cdot q \cdot |P∩Q|$ divides $|N_G(P∩Q)|$ for some prime $q$ other than $p$.

Proof: Note that since $p$-groups are nilpotent, $P∩Q < N_P(P∩Q)$ and $P∩Q < N_Q(P∩Q)$, so that $p \cdot |P∩Q|$ divides $|N_G(P∩Q)|$. Prove that $N_P(P∩Q)$ and $N_Q(P∩Q)$ are distinct Sylow $p$-subgroups of $N_G(P∩Q)$. If $N_P(P∩Q)$ is not a Sylow subgroup, then place it in one, and call this group $P^*$. This is a $p$-group in $G$, so place $P^*$ in a Sylow $p$-subgroup $P^{**}$. Note that if $P^{**}=P$, then since $P^* ≤ N_G(P∩Q)$ we have $P^*∩P = P^* ≤ N_P(P∩Q)$, a contradiction. Therefore $P$ and $P^{**}$ are distinct Sylow $p$-subgroups of $G$ and $P∩Q < N_P(P∩Q) ≤ P∩P^{**}$, a contradiction by the maximality of $P∩Q$. The case for $N_Q(P∩Q)∈Syl_p(N_G(P∩Q))$ is similar. Now prove that $N_P(P∩Q)≠N_Q(P∩Q)$: Assuming the contrary, we have $P∩Q < N_P(P∩Q) ≤ P∩Q$, a clear impossibility.

Assume $N_G(P∩Q)$ is a $p$-group, and thus place it in a Sylow $p$-subgroup of $G$, calling it $M$. If $M=P$, then $N_Q(P∩Q) ≤ P$, so that $N_Q(P∩Q) ≤ P∩Q$, untenable. Therefore $M≠P$ and $P∩Q < N_P(P∩Q) ≤ M∩P$, another contradiction. Therefore $q$ divides $|N_G(P∩Q)|$.

Finally, let $A,B∈Syl_p(N_G(P∩Q))$ with $A ≠ B$, and prove $A∩B = P∩Q$. ($\supseteq$) Let $P∩Q ≤ C$ for some Sylow $p$-subgroup of $N_G(P∩Q)$ with $gCg^{-1}=A$ for some $g∈N_G(P∩Q)$. We have $g(P∩Q)g^{-1}=P∩Q≤A$. The case is parallel for $P∩Q ≤ B$. ($\subseteq$) $A$ and $B$ are $p$-subgroups of $G$, so let $A≤D$ and $B≤E$ for some $D,E∈Syl_p(G)$. Assume $D=E$: Since $A≠B$ and $A,B≤D$, we have $\langle~A,~B~\rangle$ is a $p$-subgroup (because it is in $D$) properly containing $A$ and $B$, a contradiction (since $\langle~A,~B~\rangle ≤ N_G(P∩Q)$ and $A$ and $B$ are supposed to be Sylow in $N_G(P∩Q)$). So $D≠E$. We have $A∩B≤D∩E$ so that $|A∩B|≤|D∩E|≤|P∩Q|$ and now $A∩B=P∩Q$.$~\square$

Analysis of Groups of Order 2205 (6.2.6)

Dummit and Foote Abstract Algebra, section 6.2, exercise 6 (excerpt):

Prove there are no simple groups of order $2205=3^2 \cdot 5 \cdot 7^2$.

Proof: Assume $G$ to be a contradiction. By Sylow analysis, we have: $$n_3∈\{7,49\}~~~~~n_5∈\{21,441\}~~~~~n_7=15$$ Index considerations disallow $n_3=7$, though we need only look at $n_7=15$. Since $15 \not ≡ 1 \mod{7^2}$, set $P_0=P_1∩P_2$ for $P_1,P_2∈Syl_7(G)$ such that $|P_0|=7$. We have $P_0 \trianglelefteq P_1, P_2$ so that $7^2 \mid |N_G(P_0)| ≠ 7^2$. Due to index considerations once more, we have $|N_G(P_0)|=3 \cdot 7^2$. But in this case $n_7(N_G(P_0))=1$, so that there is a unique Sylow 7-subgroup of $N_G(P_0)$, despite $P_1$ and $P_2$ being distinct groups of order $7^2$ contained therein.$~\square$

Hall's Theorem of Solvable Groups (6.1.33)

Dummit and Foote Abstract Algebra, section 6.1, exercise 33:

Let $\pi$ be any set of primes. A subgroup $H$ of $G$ is called a Hall $\pi$-subgroup if the only primes dividing $|H|$ are in $\pi$ and $(|H|,|G:H|)=1$. When $\pi=\{p\}$, these groups are known as Sylow $p$-subgroups. Prove the following generalization of Sylow's Theorem for solvable groups: If $G$ is a finite solvable group, then for any set of primes $\pi$, the set of Hall $\pi$-subgroups is nonempty and all of its elements are conjugate to each other in $G$. [Let $M$ be a minimal normal subgroup of $G$, so $M$ is a $p$-group. If $p∈\pi$, then proceed by induction on $G/M$. If $p \not ∈\pi$, then reduce to the case where $|G|=p^an$, where $n$ is the order of a Hall $\pi$-subgroup. In this case let $N/M$ be a minimal normal $q$-subgroup of $G/M$, and let $Q∈Syl_q(N)$. If $Q \trianglelefteq G$, then proceed with $Q$ in place of $M$. If not, then use Frattini's argument to show $N_G(N)$ is a Hall $\pi$-subgroup and establish conjugacy as well.]

Proof: Let the overbar denote passage into $G/M$. If $p∈\pi$, then by induction there is a Hall $\pi$-subgroup of $\overline{G}$ of order $n/p^a$, so that its preimage in $G$ is of order $n$, and is thus a Hall $\pi$-subgroup. Furthermore, if any Hall $\pi$-subgroup $H$ does not contain $M$, then $H∩M≠M$, so that $p$ divides $| M : H∩M |$, so that by the Second Isomorphism Theorem $p$ divides $| HM : H |$, which implies $p$ divides $|G:HM| \cdot |HM : H|=|G:H|$, a contradiction. Now, for any two Hall $\pi$-subgroups $H_1$ and $H_2$, we have $\overline{H_1}$ and $\overline{H_2}$ are Hall $\pi$-subgroups of $\overline{G}$, and by induction they are conjugate, so that $\overline{g}\overline{H_1}\overline{g}^{-1}=\overline{gH_1g^{-1}}=\overline{H_2}$. Since $|gH_1g^{-1}|=n$ we have $gH_1g^{-1}$ is a Hall $\pi$-subgroup, and now both these groups contain $M$ so that $gH_1g^{-1}=H_2$.

Therefore assume $p \not ∈ \pi$. Taking the Hall $\pi$-subgroup $\overline{H}$ of $\overline{G}$ and taking its preimage $HM$ of order $p^an$, we can be assured that if $HM < G$, then by induction we can recognize Hall $\pi$-subgroups of $HM$ for the existence condition. For any two Hall $\pi$-subgroups $H_1$ and $H_2$ of $G$, since $H_1∩M=H_2∩M=1$, we have Hall $\pi$-subgroups $\overline{H_1}$ and $\overline{H_2}$ of $\overline{G}$, so that $\overline{g}\overline{H_1}\overline{g}^{-1}=\overline{H_2}$ by induction, and now $gH_1g^{-1}M=gH_1Mg^{-1}=H_2M$, so that $gH_1g^{-1}≤H_2M$. This reveals $H_1$ is conjugate to some Hall $\pi$-subgroup of $H_2M$, and by induction this one is conjugate to $H_2$, so that finally $H_1$ is conjugate to $H_2$.

So assume $HM=G$ and $|G|=p^an$. If $G$ is a $p$-group the proposition is evident, so we can take a minimal normal subgroup $\overline{N}$ (of order $q^b$ with $q≠p$) of $\overline{G}$, and observe $Q∈Syl_q(N)$. If $Q \trianglelefteq G$, then we can argue with $Q$ in place of $M$ as above. Therefore assume $N_G(Q) < G$. Since $\overline{N} \trianglelefteq \overline{G} ⇒ N \trianglelefteq G$, we can apply Frattini's Argument to $N$. Notice: $$|G|=p^an=\dfrac{|N| \cdot |N_G(Q)|}{|N∩N_G(Q)|}=\dfrac{p^aq^b \cdot |N_G(Q)|}{|N∩N_G(Q)|}$$ Since $Q≤N$ and $Q≤N_G(Q)$, we have $q^b$ divides $|N∩N_G(Q)|$. Therefore, letting $n=q^{\alpha_0}p_1^{\alpha_1}...p_r^{\alpha_r}$, we observe that $q^{\alpha_0}$ and $p_i^{\alpha_i}$ divides $|N_G(Q)|$ for all $i$, ensuring that $n$ divides $|N_G(Q)|$ and now by induction there exist Hall $\pi$-subgroups of $G$. Establish conjugacy: Take an arbitrary Hall $\pi$-subgroup $T$ of $G$. Since $MT=G$ and $M≤N$, we have: $$\dfrac{|N| \cdot |T|}{|N∩T|}=|G|⇒\dfrac{p^aq^b \cdot n}{|N∩T|}=p^an⇒|N∩T|=q^b=|Q|$$ So that $N∩T$ is a Sylow $q$-subgroup of $N$ and is thus conjugate to $Q$; let $g(N∩T)g^{-1}=Q$. Since clearly $T≤N_G(N∩T)$, we have $gTg^{-1}≤gN_G(N∩T)g^{-1}=N_G(g(N∩T)g^{-1})=N_G(Q)$, so that every Hall $\pi$-subgroup of $G$ is conjugate to some Hall $\pi$-subgroup of $N_G(Q)$, whose Hall $\pi$-subgroups are inductively assumed to be conjugate to each other.$~\square$

Minimal Normal Subgroups of Solvable Groups (6.1.31)

Dummit and Foote Abstract Algebra, section 6.1, exercise 31:

A group $M$ is called a minimal normal subgroup of $G$ when $M \trianglelefteq G$ and every proper nontrivial subgroup of $M$ is nonnormal in $G$. Prove that when $G$ is finite and solvable, any minimal subgroup $M$ is an elementary abelian $p$-group for some prime $p$. [Examine $M$'s characteristic subgroups $M'$ and $\langle~x^p~|~x∈P~\rangle$]

Proof: Let $M^p$ denote $\langle~x^p~|~x∈P~\rangle$. If $M$ is trivial, the case holds, so assume $M≠1$.

Lemma 1: $G^n~\text{char}~G$ for any finitely generated group $G$ and $n∈\mathbb{Z}$. Proof: $$x∈G^n⇔x=y_1^n...y_r^n⇔φ(x)=φ(y_1^n...y_r^n)=φ(y_1)^n...φ(y_r)^n⇔φ(x)∈G^n~\square$$

Since any characteristic subgroup of $M$ is concomitantly normal in $G$, we must not have any proper nontrivial characteristic subgroups of $M$. Assume $M'=M$; since subgroups of a solvable group are solvable by Proposition 10(1), and yet the derived chain of $M$ terminates before reaching the identity, we have a contradiction. So assume $M'=1$, and now $M$ is abelian.

Suppose $M^p=1$ for some prime $p$. Then all the generators of $M^p$ are trivial, which is to say $x^p=1$ for all $x∈M$ and now $M$ is elementary abelian. So assume that $M^p=M$ for any prime $p$. Inductively prove that $M^{n}=M$ for all $n∈\mathbb{Z^+}$, an absurd conclusion for $M^{|M|}=M$. The case is clear when $n=1$, so proceed to the inductive step. The case is assumed true for prime $n$, so we have prime $q$ dividing $n≠q$. We have $(M^{n/q})^q=M$ by induction, revealing that for any element of the abelian $M$ we have a representation of the form $(m_{1,1}^{n/q}...m_{1,r_1}^{n/q})^q(m_{2,1}^{n/q}...m_{2,r_2}^{n/q})^q...(m_{s,1}^{n/q}...m_{s,r_s}^{n/q})^q=$$m_{1,1}^n...m_{1,r_1}^nm_{2,1}^n...m_{2,r_2}^n...m_{s,1}^n...m_{s,r_s}^n∈M^n$.$~\square$

Frattini Subgroup of p-Groups (6.1.26a-b)

Dummit and Foote Abstract Algebra, section 6.1, exercise 26a-b:

(a) Let the overbar denote passage into $P/\Phi(P)$. Prove $\overline{P}$ is an elementary abelian $p$-group. [Prove $P'≤\Phi(P)$ and $x^p∈\Phi(P)$ for any $x∈P$]

(b) Prove that if $P/N$ is elementary abelian, then $\Phi(P)≤N$.

Proof: (a) Note that $p$-groups are nilpotent, so that every maximal subgroup of $P$ is normal and of prime index by 6.1.4, so that the quotient group of $P$ over any maximal subgroup is abelian, so that by an extension of 5.4.14, $P'$ is contained within the intersection of all maximal subgroups, which is to say $P'≤\Phi(P)$. Similarly, since the quotient groups are of prime order, we have $\overline{x}^p$ is trivial for any $x∈P$, which is to say $x^p∈\Phi(P)$.$~\square$

(b) Lemma 1: $\Phi(Z_p^t)=1$ for any prime $p$ and nonnegative integer $t$. Proof: Let $(x_1,...,x_r,...,x_t)$ be any nontrivial element with a nontrivial coordinate designated by $x_r$. We have $\langle~e_1,...,~e_{r-1},~e_{r+1},...,~e_t~\rangle$ is a subgroup of order $p^{t-1}$ that is maximal by its index, and this particular subgroup doesn't contain $(x_1,...,x_r,...,x_t)$. Since the nontrivial element was arbitrary, we have $\Phi(Z_p^t)=1$.$~\square$

Since $P/N$ is elementary abelian, we have the intersection of its maximal subgroups is the identity, which is to say (by the Lattice Isomorphism Theorem) that the intersection of the set of maximal subgroups of $P$ containing $N$ is contained in $N$. By definition, $N$ is contained within the intersection of the set of maximal subgroups of $P$ containing $N$, so there is an equality. Since $\Phi(P)$ is contained in all maximal subgroups of $P$, we have $\Phi(P)≤N$.

In other words, if $φ : P → E$ is any homomorphism of $P$ into an elementary abelian group $E$, then $φ$ factors through $\Phi(P)$. $$\begin{array}{ccc} P&\stackrel{\pi}{\longrightarrow}&{P/\Phi(P)}\\ &\searrow\scriptstyle{φ}&\downarrow\scriptstyle{ψ}\\ &&E \end{array}$$ $~\square$

On Frattini Subgroups (6.1.21-25)

Dummit and Foote Abstract Algebra, section 6.1, exercises 21-25:

21. Prove $\Phi(G)~\text{char}~G$.
22. Prove $N \trianglelefteq G ⇒ \Phi(N)≤\Phi(G)$. Give an explicit example where this containment does not hold when $N \not \trianglelefteq G$.
23. Compute $\Phi(S_3)$,$\Phi(A_4)$,$\Phi(S_4)$,$\Phi(A_5)$, and $\Phi(S_5)$.
24. Define $x~\text{is a nongenerator}~⇔∀H < G~[\langle~x,~H~\rangle ≠ G]$. Prove $\Phi(G)= \{x~|~x~\text{is a nongenerator}~\}$ when $G ≠ 1$.
25. Let $|G|< \infty$. Prove $\Phi(G)$ is nilpotent. [Use Frattini's Argument to show $∀P∈Syl(\Phi(G))~[P \trianglelefteq \Phi(G)]$]

Proof: Let $\mathcal{M}=\{...,H_i,...\}$ be the (potentially infinite, even uncountable) set of maximal subgroups of $G$, and $I$ be its indexing set.

Lemma 1: For any $φ∈Aut(G)$, we have $φ$ permutes the elements of $\mathcal{M}$. Proof: Pick an arbitrary $H_i∈\mathcal{M}$. If $φ(H_i)$ is not a maximal subgroup of $G$, then observe $φ(H_i) < K < G$. We have $H_i = φ^{-1}(φ(H_i)) < φ^{-1}(K)$, the last term of which is properly contained in $G$ (else $K=φ(G)=G$), therefore $H_i < φ^{-1}(K) < G$, a contradiction. Therefore $φ(H_i)$ is a maximal subgroup, and this action of $φ$ on $\mathcal{M}$ is well defined, injective by nature, and surjective since $φ^{-1}(H_i)$ will map to $H_i$. $\square$

Lemma 2: $φ(\bigcap_{i∈I}H_{i})=\bigcap_{i∈I}φ(H_i)$ for any isomorphism $φ$ of $G$, with $H_i≤G$ for all $i∈I$. Proof: $$φ(x)∈φ(\bigcap_{i∈I}H_{i})⇔x∈\bigcap_{i∈I}H_{i}⇔∀i∈I[x∈H_i]⇔$$ $$∀i∈I[φ(x)∈φ(H_i)]⇔φ(x)∈\bigcap_{i∈I}φ(H_{i})~~~\square$$

(21) Let $\pi$ denote the permutation of the indices of the elements of $\mathcal{M}$ by $φ∈Aut(G)$ by lemma 1. We have $$φ(\Phi(G))=φ(\bigcap_{i∈I}H_{i})=\bigcap_{i∈I}φ(H_{i})=\bigcap_{i∈I}H_{\pi(i)}=\bigcap_{i∈I}H_{i}=\Phi(G)$$
since $\pi(I)=I$.$~\square$

(22) (With aid from Project Crazy Project) (Assuming $G$ is finite) Lemma 3: Let $A ≤ C ≤ G$ and $B ≤ G$. Then $A(B∩C)=AB∩C$. Proof: ($\subseteq$) $x∈A(B∩C)⇒x=ad$ for some $a∈A$ and $d∈B∩C$, so that $ad∈AB$ and $ad∈C$, ergo $x∈AB∩C$. ($\supseteq$) $x∈AB∩C⇒x=ab \land x∈C$, so that $b=a^{-1}x∈C$ and now $b∈B∩C$, hence $x∈A(B∩C)$. $~\square$

Lemma 4: $\Phi(G)H < G$ for any $H < G$. Proof: Quickly done, this is a consequence of exercise 24. Proving it otherwise is a simple feat nonetheless.$~\square$

Now, we have $\Phi(N)~\text{char}~N \trianglelefteq G⇒\Phi(N) \trianglelefteq G$, so that $\Phi(N)K$ is a subgroup for any $K≤G$. Observe the implications: $$\Phi(N) \not ≤ \Phi(G )⇒ ∃i∈I[\Phi(N) \not ≤ H_i]⇒N \not ≤ H_i ⇒ H_i∩N < N ⇒$$ $$\Phi(N)(H_i∩N) < N ⇒ \Phi(N)H_i∩N < N ⇒ G∩N < N ⇒ N < N$$ We proceed to constructing a counterexample when $N \not \trianglelefteq G$. This lemma will be useful for the next exercise as well: Lemma 5: $\Phi(S_n)=\Phi(A_n)=1$ when $n \geq 5$. Proof: $\Phi(A_n) \triangleleft A_n$, so we must have $\Phi(A_n)=1$. As computed in 4.6.2, the only proper normal subgroups of $S_n$ are $A_n$ and $1$. Since $A_n$ is a maximal subgroup and not the only one (place $\langle~(1 2)~\rangle$ in a maximal subgroup, for instance), we must have $\Phi(S_n)=1$.$~\square$

Now, associate $Q_8$ with its isomorphic image in $S_8$. By order considerations, it is not equal to $S_8$,$A_8$, or $1$, so it is nonnormal. We can see $\Phi(Q_8)=\langle~-1~\rangle$ is nontrivial, and by the above lemma $\Phi(S_8)$ is trivial, so the containment doesn't hold.$~\square$

(23) The groups of order 3 and 2 by Cauchy are maximal by their orders, and have trivial intersection, so $\Phi(S_3)=1$. By 3.5.8, $\langle~(1 2)(3 4),~(1 3)(2 4)~\rangle$ and $\langle~(1 2 3)~\rangle$ are maximal subgroups, and have trivial intersection, therefore $\Phi(A_4)=1$.

For $\Phi(S_4)$, take an arbitrary subgroup $M$ of order 12 in $S_4$. By Cauchy, $M$ has an element of order 3, which must be a 3-cycle. Since conjugation of $M$ by elements of $S_4$ are automorphisms of $M$ (since $M \trianglelefteq S_4$ by order), we have that $M$ contains all 3-cycles, and is thus equal to $A_4$. Now, $\langle~(1 2),(1 2 3)~\rangle$ can only contain elements of $S_3$ due to its generators fixing 4, and indeed it does generate at least and thus exactly 6 elements. Therefore it is maximal since otherwise it is contained in a group of order 12, which must be $A_4$ by above, a contradiction since $A_4$ doesn't contain a subgroup of order 6 by 3.5.8. By this, $\Phi(S_4)$ is either of order 1 or 3, and not the latter since $\Phi(S_4) \trianglelefteq S_4$ would imply $\Phi(S_4)$ contains every 3-cycle.

By the lemma, $\Phi(A_5)=\Phi(S_5)=1$.$~\square$

(24) (Assuming $G$ is finite)($\subseteq$) Let $x∈\Phi(G)$, and for any $H < G$, place $H$ in a maximal subgroup $H^*$. We have $x∈H^*$, therefore $\langle~x,~H~\rangle ≤ \langle~x,~H^*~\rangle = H^* < G$. ($\supseteq$) Let $x$ be a nongenerator, so that for any maximal subgroup $H^*$, we must have $\langle~x,~H^*~\rangle = H^*$, so that $x∈H^*$.$~\square$

(25) Let $P$ be an arbitrary Sylow subgroup of $\Phi(G)$. By Frattini, we have $\Phi(G)N_G(P)=G$. Since $\Phi(G)N_G(P)$ is clearly not a proper subgroup, we must have $N_G(P)$ is not a proper subgroup by lemma 4, which is to say $N_G(P)=G$ and now $P \trianglelefteq G$, and in particular $P \trianglelefteq \Phi(G)$, so that $\Phi(G)$ is nilpotent by theorem 3(3).$~\square$