Category Archives: Incomplete

A flatness criterion for modules

Let $R$ be a ring with 1, and let $A$ be a right unital $R$-module.

1. Prove that if $A$ is flat then for every ideal $I \subseteq R$ the mapping $A \otimes_R I \rightarrow A \otimes_R R$ induced by the inclusion $I \rightarrow R$ is injective.
2. Suppose that for every finitely generated ideal $I \subseteq R$, the mapping $A \otimes_R I \rightarrow A \otimes_R R$ induced by the inclusion map is injective. Prove that in fact the induced map $A \otimes_R I \rightarrow A \otimes_R R$ is injective for every ideal $I$. Show further that if $K$ is any submodule of a finitely generated free module $F$ then $A \otimes_R K \rightarrow A \otimes_R F$ is injective. Show that the same is true for any free module (not necessarily finitely generated).
3. Under the hypothesis in part (2), suppose $\psi : L \rightarrow M$ is an injective left $R$-module homomorphism. Prove that the induced map $1 \otimes \psi : A \otimes_R L \rightarrow A \otimes_R M$ is injective. Conclude that $A$ is flat.
4. Let $F$ be a flat right unital $R$-module and $K \subseteq F$ an $R$-submodule of $F$. Prove that $F/K$ is flat if and only if $FI \cap K = KI$ for every finitely generated ideal $I \subseteq R$.

1. This is true by our definition of flat module.
2. Suppose the induced map $1 \otimes \iota : A \otimes_R I \rightarrow A \otimes_R R$ is injective for all finitely generated ideals $I \subseteq R$. Let $I \subseteq R$ be an arbitrary ideal and consider the induced map $1 \otimes \iota : A \otimes_R I \rightarrow A \otimes_R R$. Suppose $(1 \otimes \iota)(\sum a_i \otimes r_i) = 0$. Now there is a finitely generated ideal $J \subseteq R$ which contains all of the $r_i$. Moreover, $1 \otimes \iota$ restricted to $A \otimes J$ is injective by our hypothesis. Thus $\sum a_i \otimes r_i = 0$, and so $1 \otimes \iota$ is injective.

Now let $F = \oplus_T R$ be a finitely generated free module, and let $K \subseteq F$ be a submodule. Now $K \subseteq \oplus_T I_t$, where $I_t \subseteq R$ is a submodule (i.e. a left ideal). (Specifically, $I_t$ is the set of all $t$th coordinates of elements of $K$. This set is an $R$-submodule of $R$ (an ideal) since $K$ is an $R$-module.) Now $A \otimes_R (\oplus_T I_t)$ $\cong \oplus_T (A \otimes_R I_t)$ $\hookrightarrow \oplus_T A \otimes_R R$ $\cong A \otimes_R (\oplus_T R)$ $= A \otimes_R F$. Thus the induced map $1 \otimes \iota : A \otimes_R K \rightarrow A \otimes_R F$ is injective.

Finally, let $F$ be an arbitrary free module and let $K \subseteq F$ be an arbitrary submodule. Consider the induced map $1 \otimes \iota : A \otimes_R K \rightarrow A \otimes_R F$. If $(1 \otimes \iota)(\sum a_i \otimes k_i) = 0$, then there is a finitely generated submodule of $F$ containing all the $k_i$. Moreover, $1 \otimes \iota$ restricted to this submodule is injective, so that $\sum a_i \otimes k_i = 0$. Thus $1 \otimes \iota$ is injective.

3. Let $\psi : L \rightarrow M$ be an injective left module homomorphism. Now every module (for instance $M$) is isomorphic to a quotient of a free module. Suppose $F$ is a free module and $K \subseteq F$ a submodule such that $M = F/K$. In particular, the sequence $0 \rightarrow K \rightarrow F \rightarrow M \rightarrow 0$ is short exact. Let $J \subseteq F$ be the subset consisting of precisely those elements $x \in F$ such that $\pi(x) \in \mathsf{im}\ \psi$, where $\pi : F \rightarrow M$ is the natural projection. That is, $J = \{ x \in F \ |\ \pi(x) \in \mathsf{im}\ \psi \}$. Define $\theta : J \rightarrow L$ by $\theta(x) = y$, where $\pi(x) = \psi(y)$. Now $\theta$ is total by definition and is well-defined because $\psi$ is injective. Suppose now that $x,y \in J$ and $r \in R$. There exist $a,b \in L$ such that $\theta(x) = a$ and $\theta(y) = b$. So $\pi(x) = \psi(a)$ and $\pi(y) = \psi(b)$. Now $\pi(x) + r\pi(y) = \psi(a) + r\psi(b)$, so that $\pi(x+ry) = \psi(a+rb)$. Thus we have $\theta(x+ry) = a+rb = \theta(x) + r\theta(y)$, and so $\theta$ is a left $R$-module homomorphism. We claim also that $K \subseteq J$. It is certainly the case that $K \subseteq J$, since $K = \mathsf{ker}\ \pi$. This gives the following diagram of module homomorphisms.

A diagram of modules

We claim that this diagram commutes and that the rows are exact. First we show exactness; the bottom row is exact by definition. For the top row, certainly $\mathsf{im}\ \iota \subseteq \mathsf{ker}\ \theta$, since $\psi$ is injective. Now suppose $x \in \mathsf{ker}\ \theta$. Then $\pi(x) = 0$, so that $x \in \mathsf{ker}\ \pi = K = \mathsf{im}\ \iota$. Thus the rows are exact. Now we show that the diagram commutes. The left square certainly commutes, since the maps are all simply inclusions. Now let $x \in J$. Note that $(\psi \circ \theta)(x) = \psi(\theta(x))$ $= \psi(y)$, where $\pi(x) = \psi(y)$. Thus $\psi \circ \theta = \pi$, and so the diagram commutes. In fact, the triple $(\mathsf{id}, \iota, \psi)$ is a homomorphism of short exact sequences.

Recall that the functor $A \otimes_R -$ is right exact. Thus the induced diagram of modules

Another diagram of modules

commutes and has exact rows. By part (2) above, $1 \otimes \iota : A \otimes_R J \rightarrow A \otimes_R F$ is injective, and certainly $\mathsf{id} : A \otimes_R K \rightarrow A \otimes_R K$ and $1 \otimes \theta$ are surjective. By part 4 of this previous exercise, $1 \otimes \psi$ is injective.

Thus $A$ is flat as a right $R$-module.

4. Suppose $F$ is flat, $K \subseteq F$ a submodule, and that for all finitely generated ideals $I$ we have $FI \cap K = KI$. We wish to show that $F/K$ is flat. Let $I \subseteq R$ be an arbitrary finitely generated ideal. By the flatness criterion proved above, it suffices to show that $1 \otimes \iota : F/K \otimes_R I \rightarrow F/K \otimes_R R$ is injective. To that end, consider the exact sequences shown in the rows of the following diagram.

Yet another diagram of modules

With $t_K$ given by $k \otimes a \mapsto ka$ (and likewise $t_F$) and $\iota_K$ by $k \mapsto k \otimes 1$ (and likewise $\iota_F$), we claim that there exist suitable injective maps $\alpha$ and $\beta$ so that this diagram commutes.

Define $\alpha^\prime : F/K \times I \rightarrow FI/KI$ by $\alpha^\prime(x+K, a) = xa+KI$. Certainly if $x-y \in K$, then $xa-ya \in KI$, so that $\alpha^\prime$ is well defined. Moreover, this map is clearly bilinear. Thus we have an $R$-module homomorphism $\alpha : F/K \otimes_R I \rightarrow FI/KI$ given by $(x+K) \otimes a \mapsto xa+KI$. Clearly the upper right hand square commutes.

Note that $t_K$ is surjective (obviously). Moreover, since $F$ is flat, the map $1 \otimes \iota : F \otimes_R I \rightarrow F \otimes_R R$ is injective (by part (1) above). Now $t_F : F \otimes_R R \rightarrow F$ is an isomorphism and we have $t_F = t_F \circ (1 \otimes \iota)$, so that $t_F$ in our diagram is injective. By part 1 of this previous exercise, $\alpha$ is injective.

Now define $\beta : FI/KI \rightarrow F/K \otimes_R R$ by $x+KI \mapsto x \otimes 1$. First, if $x-y \in KI \subseteq K$, then $(x+K) \otimes 1 = (y+K) \otimes 1$, so that $\beta$ is well defined. It is clear that $\beta$ is a homomorphism of modules, and that the lower right hand square commutes. We claim also that $\beta$ is injective. To see this, suppose $x+KI \in \mathsf{ker}\ \beta$. Now $(x+K) \otimes 1 = 0$. Recall that $F/K \otimes_R R \cong F/K$ via the “multiplication” homomorphism $(x+K) \otimes r \mapsto xr+K$; so in fact we have $x+K = 0$, so that $x \in K$. Now $x \in FI$ and $x \in K$, so that (by our hypothesis) $x \in KI$. So $x+KI = 0$, and in fact $\beta$ is injective.

Finally, we claim that $\beta \circ \alpha = 1 \otimes \iota$. To see this, note that $(\beta \circ \alpha)((x+k) \otimes a) = \beta(xa+KI) = (xa+K) \otimes 1$ $= (x+K) \otimes a$.

Thus $1 \otimes \iota : F/K \otimes_R I \rightarrow F/K \otimes_R R$ is injective for all finitely generated ideals $I \subseteq R$. By the flatness criterion, $F/K$ is flat as a right $R$-module.

Conversely, suppose $F$ and $F/K$ are flat and let $I \subseteq R$ be a finitely generated ideal. Note that $KI \subseteq FI$ and $KI \subseteq K$, so that the inclusion $KI \subseteq FI \cap K$ always holds. Now consider the following commutative diagram of modules.

Last one, I promise!

Again, $t_K$ (and $t_F$ and $t_{F/K}$ denotes the isomorphism $K \otimes_R R \rightarrow K$ given by $k \otimes r \mapsto kr$. Let $\sum \zeta = x_ia_i \in FI \cap K$. (Letting $x_i \in F$ and $a_i \in I$.) We will chase $\zeta$ around the diagram to see that $\zeta \in KI$. Note that $0 = \pi(\zeta) = \sum x_ia_i + K$, so that $\sum (x_i+K) \otimes a_i = 0$ in $F/K \otimes_R R$. Note that $(1 \otimes \iota)(\sum (x_i+K) \otimes a_i) = \sum (x_i+K) \otimes a_i$, and because $1 \otimes \iota$ is injective (by our hypothesis that $F/K$ is flat), $\sum (x_i+K) \otimes a_i = 0$ in $F/K \otimes_R I$. In particular, $\sum x_i \otimes a_i \in \mathsf{ker}\ \pi \otimes 1$. Since the top row is exact, we have $\sum x_i \otimes a_i = \sum y_i \otimes b_i$ for some $y_i \in K$ and $b_i \in I$. Following these elements down to $F$, we see that $\zeta = \sum y_ib_i \in KI$ as desired.

Thus if $F$ and $F/K$ are flat, then $FI \cap K = KI$ for all finitely generated ideals $I \subseteq R$.

Subgroups of finitely generated groups need not be finitely generated

Prove that the commutator subgroup of the free group on 2 generators is not finitely generated. (In particular, subgroups of finitely generated groups need not be finitely generated.)

We begin with a lemma.

Lemma: A free group of countably infinite rank is not finitely generated. Proof: Let $A$ be a countably infinite set and suppose $F(A)$ is finitely generated, say by $S$. Now only a finite number of elements of $A$ appear as letters in the words in $S$. Thus some word in $F(A)$ is not in $\langle S \rangle$, a contradiction. $\square$

Let $F_2 = F(a,b)$ be the free group of rank 2 and let $C = \{ [x,y] \ |\ x,y \in F_2 \}$ be the set of commutators in $F_2$. By the universal property of free groups and using the inclusion $C \rightarrow F_2^\prime$, there exists a unique group homomorphism $\Phi : F(C) \rightarrow F_2^\prime$, which is surjective by construction. Now suppose $w \in \mathsf{ker}(\Phi)$. Now $\Phi(w) = 1$ is a word in $F_2^\prime \leq F_2$; since this group is free on $a$ and $b$, we must have $w = 1$. Thus $\Phi$ is injective, and we have $F_2^\prime \cong F(C)$. Hence $F_2^\prime$ is a free group of infinite rank. By the lemma, $F_2^\prime$ is not finitely generated.

A finite group whose every proper subgroup is nilpotent is solvable

Prove that if $G$ is a finite group in which every proper subgroup is nilpotent, then $G$ is solvable.

Let $\mathcal{C}$ denote the class of all finite groups $G$ such that every proper subgroup of $G$ is nilpotent and $G$ is not solvable. Choose $G \in \mathcal{C}$ of minimal order.

Suppose $G$ is not simple; say $N \leq G$ is nontrivial and normal. Note that every proper subgroup of $N$ and of $G/N$ is nilpotent, and that $N$ and $G/N$ have cardinality strictly less than that of $G$. Since $G$ is minimal in $\mathcal{C}$, $N$ and $G/N$ are solvable, hence $G$ is solvable, a contradiction.

Thus we may assume that $G$ is simple; that is, that $G$ has no normal subgroups. Suppose 1 is maximal in $G$; then $G$ has no proper subgroups and we have $G \cong Z_p$ for some prime $p$. But then $G$ is abelian, hence solvable, a contradiction. Thus no maximal subgroup of $G$ is trivial.

Let $M$ and $N$ be distinct maximal subgroups and suppose $M \cap N \neq 1$; then $N_G(M \cap N) \neq G$. Note that $M$ and $N$ are each nilpotent, so that $M \cap N < N_M(M \cap N)$ and $M \cap N < N_N(M \cap N)$ are strict. Thus $M \cap N \leq N_M(M \cap N) \cap N_N(M \cap N)$. Recall that if $H \leq G$, then $N_H(A) = N_G(A) \cap H$ for all subsets $A \subseteq G$, so that we have $M \cap N \leq N_G(M \cap N) \cap M \cap N$, hence $M \cap N = N_G(M\cap N)$, and thus $M \cap N = N_M(M \cap N)$. Since $M$ is nilpotent, by Theorem 3, $M \cap N \leq M$ is not proper and in fact $M = N$, a contradiction. Thus for all distinct maximal subgroups $M$ and $N$ of $G$, $M \cap N = 1$.

Let $M \leq G$ be maximal. Now $M$ is not normal in $G$, so that by §4.3 #23, the number of elements contained in conjugates of $M$ is at most $(|M|-1)[G:M] = |G| - |G|/|M| < |G|$. Some element of $G$ is not contained in a conjugate of $M$; since $G$ is finite, every element is contained in some maximal subgroup. Thus $G$ has at least two nonconjugate maximal subgroups. Call these $M$ and $N$.

Note that $MN = G$ and $M \cap N = 1$; moreover, no two maximal subgroups intersect nontrivially. Thus, via the proof of this previous exercise, counting the number of nonidentity elements in conjugates of $M$ and $N$, respectively, we see that $G$ contains at least $(|M|-1)[G:M] + (|N|-1)[G:N] = |G| + [|M| \cdot |N| - (|M| + |N|)]$ elements. Since $M$ and $N$ are nontrivial, $|M| \cdot |N| > |M| + |N|$, and $G$ has too many elements.

Thus no such group $G$ exists.

In a finite solvable group with no nontrivial normal subgroups relatively prime to p, the p-core contains its centralizer

Let $p$ be a prime dividing the order of the finite solvable group $G$. Assume $G$ has no nontrivial normal subgroups of order prime to $p$. Let $P = O_p(G)$ be the largest normal $p$-subgroup of $G$ (cf. this previous exercise). Prove that $C_G(P) \leq P$; i.e., that $C_G(P) = Z(P)$.

If $G$ is simple, then $G \cong Z_p$ for some prime $p$. In this case, $P = G$ (by definition), and we have $C_G(P) = Z(P) = G$.

Suppose $G$ is not simple. Then nontrivial normal subgroups of $G$ exist. Let $K \leq G$ be a minimal nontrivial normal subgroup; by this previous exercise, $K$ is an elementary abelian $q$-group for some prime $q$. By our hypothesis, in fact $q = p$. In particular, $P = O_p(G)$ is nontrivial.

Let $N = C_G(P)$, and let $\pi$ denote the set of primes dividing $|N|$ except $p$. ($p$ indeed does divide $N$ since, as a $p$-group, $Z(P)$ is nontrivial.) Note that $Z(P) \leq C_G(P)$ is normal and that $C_G(P) \leq N_G(P) = G$ is normal. Letting $H$ be a Hall $\pi$-subgroup of $N$ and $Q$ a Sylow $p$-subgroup, we have $N = QH$ by Lagrange. Note that $Z(P) \leq Q$. Suppose this inclusion is strict; that is, there are elements in $Q$ which are not also in $Z(P)$. Then these elements are not in $O_p(G)$. However, note that if $xO_p(G) \in G/O_p(G)$. (@@@)

We now consider $Z(P) H$. Since $Z(P) \leq P$, $H \leq C_G(Z(P))$. Now if $zh \in Z(P) H$, note that $(zh)H(zh)^{-1} = zhHh^{-1}z^{-1}$ $= zHz^{-1} = Hzz^{-1} = H$. Thus $H \leq Z(P) H$ is normal; since $H \leq Z(P) H \leq N$ is a Hall subgroup, it is in fact the unique subgroup of $Z(P) H$ of order $|H|$. Thus $H$ is characteristic in $Z(P) H$.

Recall that $C_G(P) \leq N_G(P) = G$ is normal, so that $H$ is normal in $G$. Since $|H|$ is prime to $p$ by construction, $H = 1$. Thus $C_G(P) = Z(P)$.

The Frattini subgroup is the set of all nongenerators

An element $x$ of a group $G$ is called a nongenerator if for all proper subgroups $H \leq G$, $\langle x, H \rangle \leq G$ is also proper. Prove that $\Phi(G) \leq G$ is the set $\mathsf{ng}(G)$ of all nongenerators in $G$.

$(\subseteq)$ [Not finished.] Let $x \in \Phi(G)$. Now let $H < G$ be a proper subgroup. If $H$ is contained in some maximal subgroup $M$, then we have $\Phi(G)H \leq M < G$, so that $\langle x, H \rangle < G$ is proper as desired. Suppose now that $H$ is not contained in any maximal subgroup of $G$. Suppose further that $\Phi(G)H = G$. (@@@)

[Old proof, assumes $G$ is finite.]
Let $x \in \Phi(G)$. By a lemma we proved for a previous theorem, if $H \leq G$ is proper then $\Phi(G)H \leq G$ is proper. Now $\langle x \rangle \leq \Phi(G)H$ and $H \leq \Phi(G)H$, so that $\langle x,H \rangle \leq \Phi(G)H < G$ is proper. Thus $x \in \mathsf{ng}(G)$ is a nongenerator.

$(\supseteq)$ Let $x \in \mathsf{ng}(G)$ be a nongenerator of $G$. If $G$ has no maximal subgroups, then $\Phi(G) = G$, and we have $\mathsf{ng}(G) \subseteq \Phi(G)$. Suppose now that $M \leq G$ is maximal; in particular, $M \leq G$ is proper, so that $\langle x, M \rangle \leq G$ is proper. Thus $\langle x, M \rangle \leq M$, so that $x \in M$. Since $x$ is contained in every maximal subgroup, $x \in \Phi(G)$.

Exhibit the set of all invertible diagonal matrices with diagonal entries equal as a direct product

Let $G = \{ [a_{i,j}] \in GL_n(F) \ |\ a_{i,j} = 0\ \mathrm{if}\ i > j, a_{i,i} = a_{j,j}\ \mathrm{for\ all}\ i,j \}$, where $F$ is a field, be the group of upper triangular matrices all of whose diagonal entries are equal. Prove that $G \cong D \times U$, where $D$ is the group of nonzero multiples of the identity matrix and $U$ the group of strictly upper triangular matrices.

Recall that $\overline{SL}_n(F)$ is normal in $GL_n(F)$, where $\overline{SL}_n(F)$ denotes the upper triangular matrices with only 1 on the diagonal. Then $U = G \cap \overline{SL}_n(F)$ is normal in $G$. Now if $A \in D$, we have $A = kI$ for some nonzero field element $k$. Then for all $M \in GL_n(F)$, we have $AM = kIM = kM = Mk = MkI = MA$, so that $D \leq Z(GL_n(F))$; more specifically, $D$ is normal in $G$.

We can see that $D \cap U$ is trivial, since every element in $U$ has only 1s on the main diagonal. Finally, if $A \in G$ has the element $k$ on the main diagonal, then $k^{-1}A \in U$, and in fact $A = kI \cdot k^{-1}A \in DU$. Thus $G = DU$; by the recognition theorem, we have $G \cong D \times U$.

Properties of the image and kernel of the p-power map on a finite abelian group

Let $A$ be a finite abelian group (written multiplicatively) and let $p$ be a prime. Let $A^p = \{ a^p \ |\ a \in A \}$ and $A_p = \{ x \in A \ |\ x^p = 1 \}$. (I.e. $A^p$ and $A_p$ are the image and kernel of the $p$-power map, respectively.

1. Prove that $A/A^p \cong A_p$. (Show that they are both elementary abelian and have the same order.)
2. Prove that the number of subgroups of $A$ of order $p$ equals the number of subgroups of index $p$. (Reduce to the case where $A$ is an elementary abelian $p$-group.)

1. We begin with some lemmas.

Lemma 1: Let $A$ and $B$ be abelian groups and let $\theta : A \rightarrow B$ be an isomorphism. Then there exists an isomorphism $\theta^\prime : A/A^p \rightarrow B/B^p$. Proof: Let $\pi$ denote the natural projection from $Z$ to $Z/Z^p$ (for $Z \in \{A,B\}$). If $x \in A^p$, then $x = y^p$ for some $y \in A$. Then $(\pi \circ \theta)(x) = (\pi \circ \theta)(y^p)$ $= \pi(\theta(y)^p)$ $= B^p$, so that $A^p \leq \mathsf{ker}\ (\pi \circ \theta)$. By the remarks on page 100 in D&F, there exists a unique group homomorphism $\theta^\prime : A/A^p \rightarrow B/B^p$ such that $\theta^\prime \circ \pi = \pi \circ \theta$. ($\theta^\prime$ is injective) Suppose $xA^p \in \mathsf{ker}\ \theta^\prime$. Then $\theta^\prime(\pi(x)) = 1$, thus $(\theta^\prime \circ \pi)(x) = 1$, so that $(\pi \circ \theta)(x) = 1$, hence $\theta(x) \in B^p$. Now since $\theta$ is surjective, $\theta(x) = \theta(y)^p$ for some $y \in A$. Thus $x = y^p$, so that $x \in A^p$. Thus $\mathsf{ker}\ \theta^\prime = 1$, so that $\theta^\prime$ is injective. ($\theta^\prime$ is surjective) Let $xB^p \in B/B^p$. Now $xB^p = (\pi \circ \theta)(y)$ for some $y \in A$, so that $xB^p = \theta^\prime(\pi(y))$. Hence $\theta^\prime$ is surjective. $\square$

Lemma 2: Let $A$ and $B$ be abelian groups and let $\theta : A \rightarrow B$ be an isomorphism. Then $\theta[A_p] = B_p$. Proof: $(\subseteq)$ If $x \in \theta[A_p]$, then $x = \theta(y)$ where $y \in A_p$. Now $y^p = 1$, so that $\theta(y)^p = 1$, hence $x^p = 1$. Thus $x \in B_p$. $(\supseteq)$ Let $x \in B_p$. Then x^p = 1\$. Since $\theta$ is surjective, $x = \theta(y)$ for some $y \in A$. Now $\theta(y^p) = \theta(y)^p = x^p = 1$, and since $\theta$ is injective, $y^p = 1$. Thus $y \in A_p$, hence $x \in \theta[A_p]$. $\square$

Lemma 3: Let $G$ be a finite abelian group of order $n$, let $m$ be an integer, and let $\varphi : G \rightarrow G$ denote the $m$-power homomorphism. If $\mathsf{gcd}(m,n) = 1$, then $\varphi$ is an isomorphism. Proof: Since $G$ is finite, it suffices to show that $\varphi$ is injective. Let $x \in \mathsf{ker}\ \varphi$. Then $x^m = 1$, so that $|x|$ divides $m$. By Lagrange, $|x|$ divides $n$. Thus $|x| = 1$, hence $x = 1$, and $\varphi$ is injective. $\square$

Now to the main result. By Theorem 5 in the text, $A \cong A_1 \times A_2$ where $A_1$ is a $p$-group of rank $t$ and $p$ does not divide $|A_2|$. By a lemma to the previous exercise, $\varphi = \varphi_1 \times \varphi_2$, where each map is the $p$-power map on $A_1 \times A_2$, $A_1$, and $A_2$, respectively.

We have $\mathsf{ker}\ \varphi = \mathsf{ker}(\varphi_1 \times \varphi_2)$ $= (\mathsf{ker}\ \varphi_1) \times (\mathsf{ker}\ \varphi_2)$ $\cong Z_{p^t} \times 1$ $\cong Z_{p^t}$ by the previous exercise.

Similarly, $(A_1 \times A_2)/\mathsf{im}\ \varphi = (A_1 \times A_2)/(\mathsf{im}\ \varphi_1 \times \mathsf{im}\ \varphi_2)$ $\cong (A_1/\mathsf{im}\ \varphi_1) \times (A_2/\mathsf{im}\ \varphi_2)$ $\cong Z_{p^t} \times 1$ $\cong Z_{p^t}$, again using the previous exercise.

Now the first conclusion follows using Lemmas 1 and 2.

2. We begin with some more lemmas.

Lemma 4: Let $E_{p^k}$ be the elementary abelian group of order $p^k$, and suppose $E_{p^k} = \langle S \rangle$ where $|S| = k$. Finally let $G$ be a group. If $\overline{\varphi} : S \rightarrow G$ is a mapping such that $|\overline{\varphi}(s)| = p$ for each $s \in S$ and $\overline{\varphi}(s) \overline{\varphi}(t) = \overline{\varphi}(t) \overline{\varphi}(s)$ for all $s,t \in S$, then $\overline{\varphi}$ extends uniquely to a group homomorphism $\varphi : E_{p^k} \rightarrow G$. Proof: Every element of $E_{p^k}$ can be written uniquely as $x = s_1^{a_1}s_2^{a_2} \cdots s_k^{a_k}$, where $s_1 \in S$ and $0 \leq a_k < p$. Define $\varphi(x) = \overline{\varphi}(s_1)^{a_1}\overline{\varphi}(s_2)^{a_2} \cdots \overline{\varphi}(s_k)^{a_k}$. $\varphi$ is well defined since the $S$-expansion of $x$ is unique, and is a homomorphism since the $\overline{\varphi}(s)$ commute with one another. To see uniqueness, suppose $\psi : E_{p^k} \rightarrow G$ is a group homomorphism such that $\psi(s) = \varphi(s)$ for all $s \in S$. Then $\psi(x) = \psi(s_1^{a_1}s_2^{a_2} \cdots s_k^{a_k})$ $= \psi(s_1)^{a_1} \psi(s_2)^{a_2} \cdots \psi(s_k)^{a_k}$ $= \overline{\varphi}(s_1)^{a_1}\overline{\varphi}(s_2)^{a_2} \cdots \overline{\varphi}(s_k)^{a_k}$ $= \varphi(x)$. Thus $\varphi$ is unique. $\square$

Note that every subgroup of $A$ of order $p$ is contained in $\mathsf{ker}\ \varphi$, so that in fact the order $p$ subgroups of $A$ and $\mathsf{ker}\ \varphi$ coincide. Now because $\mathsf{ker}\ \varphi \cong Z_{p^t}$ is elementary abelian, every nonidentity element has order $p$ and thus generates an order $p$ subgroup. Each such subgroup is generated by $p-1$ elements; thus there are $(p^t-1)/(p-1)$ order $p$ subgroups in $\mathsf{ker}\ \varphi$, thus in $A$.

Let $B \leq A$ be a subgroup of index $p$. Now let $x \in A^p$, say $x = y^p$, and suppose $x \notin B$. Then $y \notin B$, so that $A = \langle y \rangle B$. Now $A/B$ is a group of order $p$, so that $(yB)^p = xB = B$; then $x \in B$, a contradiction. Thus $A^p \leq B$. Moreover, by the Third Isomorphism Theorem, $[A/A^p : B/A^p] = [A:B] = p$. By the Lattice Isomorphism Theorem, the index $p$ subgroups of $A/A^p \cong Z_p^t$ correspond precisely to the index $p$ subgroups of $A$. In particular, it suffices to count the order and index $p$ subgroups in an elementary abelian $p$-group.

Now $\mathsf{Aut}(Z_p^t)$ acts on $\mathcal{M} = \{ M \leq Z_p^t \ |\ [Z_p^t : M] = p \}$ by $\varphi \cdot M = \varphi[M]$. We claim that this action is transitive. To that end, let $M_1, M_2 \leq Z_p^t$ be index $p$ subgroups. Now $M_1 \cong M_2 \cong Z_p^{t-1}$ since $M_1$ and $M_2$ are elementary abelian, thus $M_1 = \langle S_1 \rangle$ and $M_2 = \langle S_2 \rangle$ where $|S_1| = |S_2| = t-1$. Let $\overline{\psi} : S_1 \rightarrow S_2$ be a bijection, and choose some $t_1 \in Z_p^t \setminus M_1$ and $t_2 \in Z_p^t \setminus M_2$. Now $\overline{\psi} \cup \{(t_1,t_2)\}$ extends by Lemma 4 to a homomorphism $\psi : Z_p^t \rightarrow Z_p^t$. Clearly $\psi$ is surjective, hence an isomorphism, so that $\psi \in \mathsf{Aut}(Z_p^t)$. Moreover, we see that $\psi[M_1] = M_2$. Thus this action is transitive, and we have $\mathsf{Aut}(Z_p^t) \cdot M = \mathcal{M}$ for all index $p$ subgroups $M$.

Now let $M \leq Z_p^t$ be an arbitrary index $p$ subgroup and let $\varphi$ be an automorphism of $Z_p^t$ which stabilizes $M$. As above, $M = \langle S \rangle$ for some set $S$ with $|S| = t-1$. Moreover, if $x \in Z_p^t \setminus M$ then $Z_p^t = \langle S \cup \{x\} \rangle$ since $M$ is maximal. By Lemma 4, $\varphi$ is determined uniquely by is action on $S$ and $x$. We see also that since $\varphi$ stabilizes $M$, we must have $\varphi(s) \in M$ for each $s \in S$. Since $M \cong Z_p^{t-1}$ and $\varphi|_M \in \mathsf{Aut}(M)$, there are $\prod_{i=0}^{t-2}(p^{t-1} - p^{t-2-i})$ distinct choices we can make for $\varphi[S]$. Moreover, $\varphi(x)$ may be chosen arbitrarily so long as $\varphi(x) \notin M$; thus there are $p^t - p^{t-1} = p^{t-1}(p-1)$ choices for $\varphi(x)$. Thus in total $\mathsf{Stab}(M)$ contains $(p-1)p^{t-1} \prod_{i=0}^{t-2}(p^{t-1} - p^{t-2-i})$ elements.

By the Orbit-Stabilizer Theorem and Lagrange's Theorem, $|\mathcal{M}| = (\prod_{i=0}^{t-1} (p^t - p^{t-1-i}))/((p-1)p^{t-1} \prod_{i=0}^{t-2}(p^{t-1} - p^{t-2-i}))$ $= (\prod_{i=1}^t (p^t - p^{t-i}))/((p-1) \prod_{i=1}^{t-1}(p^{t} - p^{t-i}))$ $= (p^t-1)/(p-1)$.

Thus the theorem is proved.

A fact regarding conjugate normal subsets of a Sylow subgroup in a finite group

Prove that if $U$ and $W$ are normal subsets of a Sylow $p$-subgroup $P$ of a finite group $G$ then $U$ is $G$-conjugate to $W$ if and only if $U$ is $N_G(P)$-conjugate to $W$. Deduce that two elements in the center of $P$ are conjugate in $G$ if and only if they are conjugate in $N_G(P)$. (A subset $U$ is normal in $G$ if $N_G(U) = G$.)

Let $U,W \subseteq P$ with $N_P(U) = P$ and $N_P(W) = P$.

The $(\Leftarrow)$ direction is immediate. To see $(\Rightarrow)$, suppose $U$ is $G$-conjugate to $W$; then there exists $g \in G$ such that $g^{-1}Ug = W$ and $gWg^{-1} = U$. Note that $N_P(g^{-1}Ug) = g^{-1}N_{gPg^{-1}}(U)g$. Now $N_G(P) = N_G(N_P(W))$ $N_G(N_P(g^{-1}Ug))$ $= N_G(g^{-1}N_{gPg^{-1}}(U)g)$ $= g^{-1}N_G(N_{gPg^{-1}}(U))g$. (@@@)

Now suppose $x,y \in Z(P)$. In this case, $N_P(\{x\}) = C_P(\{x\}) = P$ and $N_P(\{y\}) = C_P(\{y\}) = P$, so that $\{x\}$ and $\{y\}$ are normal in $P$. By the above argument we have that $x$ and $y$ are $G$-conjugate if and only if they are $N_G(P)$-conjugate.

Every nontrivial subgroup of a solvable group contains normal abelian subgroups

Prove that if $H$ is a nontrivial subgroup of a solvable group $G$ then there is a nontrivial subgroup $A \leq H$ which is abelian and normal in $G$.

We proceed by induction on the width of $G$– that is, the number of prime factors of $|G|$ (including multiplicity).

If $G$ has width 1, then $G$ is simple and no such $H$ exists.

If $G$ has width 2, say $|G| = pq$, and $H \leq G$ is nontrivial, then without loss of generality $|H| = p$.

Lemma: Let $(\mathcal{P},\leq)$ be a nonempty partially ordered set such that every chain has bounded length. Then there exists a minimal element $m \in \mathcal{P}$. Proof: Let $\{C_i\}_{i=1}^n \subseteq \mathcal{P}$ be a chain of maximal length $n$, where $C_i \leq C_{i+1}$. If an element $C_0 \in \mathcal{P}$ exists such that $C_0 \leq C_1$, then $\{ C_i \}_{i=0}^n$ is a chain of length $n+1$, a contradiction. So $C_1$ is $\leq$-minimal in $\mathcal{P}$. $\square$

Consider the set $\mathcal{H} = \{ A \vartriangleleft G \ |\ A \leq H, A \neq 1 \}$. Note that $\mathcal{H}$ is nonempty since $H \in \mathcal{H}$. Note that because $G$ is solvable, $G$ has a finite composition series, so that by the Jordan-Holder Theorem, all composition series for $G$ have the same finite length $n$. Now every chain in $\mathcal{H}$ can be extended to a composition series for $G$, and thus has length bounded by $n$. By the lemma, then, there exists a $\leq$-minimal element $A \in \mathcal{H}$. By Lemma 5 to this previous theorem, $A$ is abelian.

Subgroups and quotient groups of solvable groups are solvable

Prove that subgroups and quotient groups of solvable groups are solvable.

We prove some lemmas first.

Lemma 1: Let $G$ be a group and let $H,K,N \leq G$ with $N$ normal in $H$. Then $N \cap K$ is normal in $H \cap K$. Proof: Let $a \in H \cap K$. Then $a(N \cap K) = aN \cap aK = Na \cap Ka = (N \cap K)a$ because $a \in H$ and $a \in K$. $\square$

Lemma 2: Let $A$, $B$, and $C$ be groups with $B$ normal in $A$. Then $B \cap C$ is normal in $A \cap C$ and there is an injective homomorphism $\varphi : (A \cap C)/(B \cap C) \rightarrow A/B$. Proof: Let $\pi : A \cap C \rightarrow A/B$ be the natural projection. Certainly $B \cap C \subseteq \mathsf{ker}\ \pi$, and if $x \in \mathsf{ker}\ \pi$, we have $x \in B$, and so $x \in B \cap C$. By the First Isomorphism Theorem, the induced homomorphism $\varphi : (A \cap C)/(B \cap C) \rightarrow A/B$ is injective. $\square$

Lemma 3: Let $G$ be a group and $H,K,N \leq G$ such that $N$ is normal in $G$ and $K \leq H$ is normal, and $N \leq K \leq H \leq G$. Then $K/N \leq H/N$ is normal. Proof: Let $a \in H$. Then $(aN)(K/N) = aK/N = Ka/N = (K/N)(aN)$. $\square$

Let $G$ be a solvable group. Then there exists a subnormal series $1 = H_1 \vartriangleleft H_2 \vartriangleleft \cdots \vartriangleleft H_{k-1} \vartriangleleft H_k = G$ such that $H_i/H_{i-1}$ is abelian for all $1 < i \leq k$.

Let $K \leq G$ be a subgroup. By Lemma 1 we have $1 = H_1 \cap K \vartriangleleft H_2 \cap K \vartriangleleft \cdots \vartriangleleft H_{n-1} \cap K \vartriangleleft H_n \cap K = K$. Moreover, by Lemma 2 we have $(H_i \cap K)/(H_{i-1} \cap K) \leq H_i/H_{i-1}$ abelian. Thus $K$ is solvable.

Let $N \leq G$ be a normal subgroup. Now $H_{i+1}N/N \leq H_iN/N$ is normal by Lemma 3, so that $1 = H_1N/N \vartriangleleft H_2N/N \vartriangleleft \cdots \vartriangleleft H_{k-1}N/N \vartriangleleft H_kN/N = G/N$. Moreover, $(H_iN/N)/(H_{i-1}N/N) \cong H_iN/H_{i-1}N$ is abelian by the Third Isomorphism Theorem. Hence $G/N$ is solvable.