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 tth 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.