Monthly Archives: June 2010

Equivalent characterizations of nilpotent and cyclic groups

If G is finite, prove that (1) G is nilpotent if and only if it has a normal subgroup of each order dividing |G|, and (2) is cyclic if and only if it has a unique subgroup of each order dividing |G|.


We begin with some lemmas.

Lemma 1: Let G be a group, and let H_1, \ldots, H_k be characteristic subgroups of G. Then H = H_1\cdots H_k is characteristic in G. Proof: H is a subgroup because each H_i is normal. If \alpha is an automorphism of G, then \alpha[H] = \alpha[H_1] \cdots \alpha[H_k] = H_1 \cdots H_k = H. \square

Lemma 2: Let G be a finite group. If P \leq G is a normal Sylow subgroup, then P is characteristic. Proof: P is the unique subgroup of order |P|. \square

Now to the main result; first we prove (1).

(\Rightarrow) We proceed by induction on the breadth (number of prime divisors with multiplicity) of G.

For the base case, if G is a finite nilpotent group of breadth 1, then G \cong Z_p is an abelian simple group. Thus G has a normal subgroup of order 1 and p, the only divisors of |G|.

For the inductive step, suppose that every finite nilpotent group of breadth k \geq 1 has a normal subgroup of each order dividing |G|, and let G be a finite nilpotent group of breadth k+1. Now Z(G) is nontrivial, so there exists an element x \in Z(G) of prime order, and \langle x \rangle G is normal. We showed in a lemma to the previous exercise that G/\langle x \rangle is nilpotent, and has breadth k. By the inductive hypothesis, G/\langle x \rangle has a normal subgroup of each order d dividing |G/\langle x \rangle|. By the Lattice Isomorphism Theorem, a subgroup \overline{H} \leq G/\langle x \rangle is normal if and only if \overline{H} = H/\langle x \rangle for some normal subgroup H \leq G. Thus we see that G has a normal subgroup of order d for all d dividing |G| such that p|d. Now let d divide |G|, with p not dividing d, and let H \leq G be a normal subgroup of order pd. Since H \leq G, H is a finite nilpotent group. By Theorem 3, H is the direct product of its Sylow subgroups. Write H = P \times Q, where P is the Sylow p subgroup of H and Q the direct product of the remaining Sylow subgroups. By Lemma 2, each factor of Q is characteristic in H, so that Q is characteristic in H. Note that |Q| = d, thus Q \leq G is normal and has order d.

(\Leftarrow) Suppose G has a normal subgroup of each order d dividing |G|. In particular, G has a normal (thus unique) Sylow p-subgroup for each prime p. By Theorem 6.3, G is nilpotent.

Now we prove (2).

The (\Rightarrow) direction was proved in Theorem 2.7.

(\Leftarrow) Suppose G is finite and has a unique subgroup of order d for each d properly dividing |G|. In particular, all the Sylow subgroups of G are unique, hence normal; say there are k of these. Suppose some Sylow p-subgroup P of G is not cyclic; then by this previous exercise, P has a subgroup congruent to Z_p \times Z_p, and thus G has distinct subgroups of order p, a contradiction. Thus each Sylow subgroup of G is cyclic; say P_i = \langle x_i \rangle. Then |x_1 \cdots x_k| = |G|, so that G = \langle x_1 \cdots x_k \rangle is cyclic.

Basic properties of finite nilpotent groups

Let G be a finite nilpotent group.

  1. If H is a nontrivial normal subgroup of G, then H intersects the center of G nontrivially. In particular, every normal subgroup of prime order is central.
  2. If H < G is a proper subgroup then H is properly contained in N_G(H).

First we prove some lemmas.

Lemma 1: Let G be a group, with A_1,A_2,B_1,B_2 \subseteq G. If A_1 \subseteq A_2 and B_1 \subseteq B_2, then [A_1,B_1] \leq [A_2,B_2]. Proof: [A_1,B_1] is generated by [a,b] such that a \in A_1 and b \in B_1. Each of these generators is in [A_2,B_2]. \square

Lemma 2: Let G be a group. If H \leq G, then H^k \leq G^k for all k, where G^k denotes the k-th term in the lower central series of G. Proof: We proceed by induction on k. For the base case, if k=0 then H^0 \leq G^0. For the inductive step, suppose H^k \leq G^k for some k \geq 0. Then by Lemma 1, H^{k+1} = [H,H^k] \leq [G,G^k] = G^{k+1}. \square

Lemma 3: If G is a nilpotent group of nilpotence class k and H \leq G, then H is nilpotent with nilpotence class at most k. Proof: By Theorem 8 in the text, G^k = 1. By Lemma 2, H^k = 1, so that H is nilpotent and of nilpotence class at most k. \square

Lemma 4: Let \varphi : G \rightarrow H be a group homomorphism and let S \subseteq G. Then \varphi[\langle S \rangle] = \langle \varphi[S] \rangle. Proof: We have x \in \varphi[\langle S \rangle] if and only if x = \varphi(s_1^{a_1} \cdots s_k^{a_k}) for some s_i \in S if and only if x = \varphi(s_1)^{a_1} \cdots \varphi(s_k)^{a_k} for some s_i \in S if and only if x \in \langle \varphi[S] \rangle. \square

Lemma 5: Let \varphi : G \rightarrow H be a surjective group homomorphism. Then H^k \leq \varphi[G^k] for all k, where G^k denotes the k-th term in the lower central series of G. Proof: We proceed by induction on k. For the base case, if k = 0 then H^0 = H = \varphi[G] = \varphi[G^0]. For the inductive step, suppose H^k \leq \varphi[G^k]. Then we have H^{k+1} = [H,H^k] \leq [\varphi[G],\varphi[G^k]] \leq \varphi[[G,G^k]] \leq \varphi[G^{k+1}]. \square

Lemma 6: Let G be a nilpotent group of nilpotence class k. If H \leq G is normal, then G/H is nilpotent of nilpotence class at most k. Proof: By Lemma 5, we have (G/H)^k \leq \varphi[G^k] = 1. \square

Now to the main results; first we show part 2. [With hints from MathReference.]

(2) Let G be a finite nilpotent group such that Z(G) \neq 1. If H \leq G is a nontrivial normal subgroup of G, then H \cap Z(G) is nontrivial. In particular, every normal subgroup of order p is contained in the center.

We proceed by induction on the breadth of G. (Recall: the breadth of a finite group is the number of prime factors dividing its order, including multiplicity.)

For the base case, if G has breadth 1, then G \cong Z_p for some prime p and thus G is simple and abelian. Thus H = G, and H \cap Z(G) \neq 1.

For the inductive step, suppose that the conclusion holds for every finite nilpotent group of breadth at most k \geq 1, and let G be a finite nilpotent group of breadth k+1. Let H \leq G be normal and suppose H \cap Z(G) = 1. Now H \cdot Z(G) \cong H \times Z(G) \leq G is an internal direct product. Moreover, since H and Z(G) are normal in G, H \cdot Z(G) \leq G is normal. Now by the Lattice Isomorphism Theorem, HZ(G)/Z(G) \leq G/Z(G) is normal. By Lemma 6, G/Z(G) is nilpotent, has breadth at most k, and HZ(G)/Z(G) \leq G/Z(G) is normal. By the inductive hypothesis, there exists an element xZ(G) \in Z(G/Z(G)) such that xZ(G) \in HZ(G)/Z(G). We can write x = hz, where h \in H and z \in Z(G), so that in fact hZ(G) \in Z(G/Z(G)). Now let g \in G be arbitrary; since H is normal in G, ghg^{-1} \in H. Now (ghg^{-1})Z(G) = (gZ(G))(hZ(G))(g^{-1}Z(G)) = hZ(G) since hZ(G) is in the center of G/Z(G), so that ghg^{-1}Z(G) = hZ(G). Write ghg^{-1} = hk, where k \in Z(G). Now ghg^{-1}h^{-1} = k \in Z(G), and since H \cap Z(G) = 1, we have ghg^{-1}h^{-1} = 1, thus gh = hg. So h \in H is in the center of G and h \neq 1, a contradiction. Thus H \cap Z(G) is nontrivial. \square

Next, we prove some more lemmas.

Lemma 7: Let G be a group, K, H_1, H_2 \leq G be subgroups with K \leq G normal, K \leq H_1, and K \leq H_2. If H_1/K = H_2/K, then H_1 = H_2. Proof: Let x \in H_1. Then xK \in H_1/K = H_2/K, so that xK = yK. In particular, x \in yK \leq H_2. The other direction is similar. \square

Lemma 8: Let G be a group, H \leq G a normal subgroup, and K a subgroup with H \leq K \leq G. Then N_{G/H}(K/H) = N_G(K)/H. Proof: Note that xH \in N_{G/H}(K/H) if and only if (xH)(K/H)(x^{-1}H) = K/H, if and only if (xKx^{-1})/H = K/H, if and only if (by Lemma 7) xKx^{-1} = K, if and only if x \in N_G(K), if and only if xH \in N_G(K)/H. \square

Now we show part 4.

(4) Let G be a finite nilpotent group. If H < G is a proper subgroup, then H < N_G(H) is proper.

Proof: We proceed again by induction on the breadth of G.

For the base case, if G has breadth 1, then G \cong Z_p is abelian. Then if H \leq G is proper, H \leq N_G(H) = G is proper.

For the inductive step, suppose the conclusion holds for all finite nilpotent groups of breadth at most k. Let G be a finite nilpotent group of breadth k+1 and let H \leq G be a proper subgroup. Now Z(G) \leq N_G(H). If Z(G) \not\leq H, then H \leq \langle H, Z(G) \rangle is proper, so that H \leq N_G(H) is proper. If Z(G) \leq H, then H/Z(G) \leq G/Z(G) is proper. Moreover, G/Z(G) is a finite nilpotent group of breadth at most k, so that H/Z(G) \leq N_{G/Z(G)}(H/Z(G)) is proper. By Lemma 8, H/Z(G) \leq N_G(H)/Z(G) is proper, so that H \leq N_G(H) is proper. \square

The higher centers in a group are characteristic

Prove that Z_k(G) is characteristic in G for all i.


We begin with a lemma.

Lemma 1: Let G be a group, let K \leq G be characteristic, and let \varphi be an automorphism of G. Then the mapping \psi : G/K \rightarrow G/K given by \psi(aK) = \varphi(a)K is an automorphism of G/K. Proof: (Well-defined) Suppose aK = bK; then ab^{-1} \in K. Since K is characteristic, \varphi(ab^{-1}) = \varphi(a)\varphi(b)^{-1} \in K, so that \varphi(a)K = \varphi(b)K. Thus \psi(aK) = \psi(bK). (Homomorphism) We have \psi((ab)K) = \varphi(ab)K = (\varphi(a)K)(\varphi(b)K) = \psi(aK)(\psi(bK). (Injective) Suppose \psi(aK) = \psi(bK). Then \varphi(ab^{-1}) \in K, so that (since K is characteristic) aK = bK. (Surjective) Let aK \in G/K. Since \varphi is surjective, aK = \varphi(b)K = \psi(bK) for some b \in G. \square

Lemma 2: Let G be a group. Suppose K \leq H \leq G such that K \leq G is characteristic and H/K \leq G/K is characteristic. Then H \leq G is characteristic. Proof: Let \varphi \in \mathsf{Aut}(G). By Lemma 1, the mapping \psi : G/K \rightarrow G/K given by \psi(aK) = \varphi(a)K is an automorphism of G/K. Since H/K is characteristic, we have \psi[H/K] = H/K; that is, if h \in H, then \psi(hK) = \varphi(h)K = xK for some x \in H. Thus \varphi(h)x^{-1} \in K, and we have \varphi(h) \in H. Thus \varphi[H] \leq H. Now suppose h \in H. Since H/K is characteristic in G/K, we have hK = \psi(xK) = \varphi(x)K for some x \in H. Then \varphi(x^{-1})h = k \in K. Since K is characteristic in G, we have k = \varphi(y) for some y \in K. Thus h = \varphi(xy), where xy \in HK = H. Thus we have H \leq \varphi[H]. Hence H \leq G is characteristic. \square

We proceed by induction on k.

For the base case, if k=0, then Z_k(G) = 1. Clearly then Z_k(G) is characteristic in G.

For the inductive step, let k \geq 0 and suppose that Z_k(G) \leq G is characteristic. Consider Z_{k+1}(G) \leq G. By definition, Z_{k+1}(G) is the subgroup of G containing Z_k(G) such that Z_{k+1}(G)/Z_k(G) = Z(G/Z_k(G)). Now by definition, Z_{k+1}(G)/Z_k(G) is characteristic in G/Z_k(G). By Lemma 2, Z_{k+1} is characteristic in G.

By induction, Z_k(G) is characteristic in G for all k.

Compute the isomorphism type of the Heisenberg group over ZZ/(p)

Let H(\mathbb{F}_p) be the Heisenberg group over \mathbb{F}_p = \mathbb{Z}/(p). Prove that H(\mathbb{F}_2) \cong D_8, and that H(\mathbb{F}_p) has exponent p and is isomorphic to the first nonabelian group in Example 7 in the text.


Define A,B \in H(\mathbb{F}_2) by A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} and B = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}. Evidently A^{-1} = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} and B^{-1} = B. Moreover, B \notin \langle A \rangle, A^4 = B^2 = 1, and AB = BA^{-1}. By Lemma 1 to a previous exercise, we have an injective group homomorphism \varphi : D_8 \rightarrow H(\mathbb{F}_2) such that \varphi(r) = A and \varphi(s) = B. Since these groups are finite and have the same cardinality, \varphi is an isomorphism.

Note that H(\mathbb{F}_p) is a nonabelian group of order p^3; thus it is isomorphic to either (Z_p \times Z_p) \rtimes Z_p or Z_{p^2} \rtimes Z_p. Now in the classification of groups of order p^3, we saw that (Z_p \times Z_p) \rtimes Z_p has no elements of order p^2.

Now in a previous exercise we showed (over \mathbb{R}, but the proof holds over any field) that for all matrices A = \begin{bmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix} in H(\mathbb{F}_p), A^p = \begin{bmatrix} 1 &  pa & pb + p \frac{p+1}{2} ac \\ 0 & 1 & pc \\ 0 & 0 & 1 \end{bmatrix}. Note that (p+1)/2 is an integer; thus A^p = 1. So H(\mathbb{F}_p) has exponent p, and thus has no elements of order p^2. Hence H(\mathbb{F}_p) \cong (Z_p \times Z_p) \rtimes Z_p.

Characterize the natural numbers which are only the orders of abelian groups

Let n > 1 be an integer. Prove the following classification: every group of order n is abelian if and only if n = p_1^{a_1} \cdots p_k^{a_k}, p_i are distinct primes, a_i \in \{1,2\}, and p_i does not divide p_j^{a_j}-1 for all i and j.


(\Rightarrow) Suppose n has the properties above; we proceed by induction on k.

For the base case, let k = 1 and let G be a group of order n = p_1^{a_1}. Then G is either cyclic or elementary abelian, hence G is abelian.

For the inductive step, suppose that every group of width at most k which satisfies the given conditions for |G| is abelian. Let G be a group of width k+1 such that |G| satisfies the given properties. Now every proper subgroup of G has width at most k, so that by the induction hypothesis, every proper subgroup of G is abelian. By this previous exercise, G is solvable.

By this previous exercise, there is a (maximal) subgroup H \leq G such that H \leq G is normal and G/H has prime order; say G/H \cong Z_p. Let P be a Sylow p-subgroup of G. Now by FTFGAG, H is the direct product of its Sylow subgroups, and the Sylow subgroups of H are normal in H. In particular, H is generated by elements of prime and prime squared order. Let x \in H have prime or prime squared order and let y \in P. If \langle xy \rangle = G, then G is cyclic, hence abelian. If \langle xy \rangle is a proper subgroup of G, then xy = yx, so that \langle x \rangle \leq C_G(y). In either case, we have \langle x \rangle \leq C_G(P). Thus H \leq C_G(P), and we have HP = G \leq C_G(P). Thus P \leq Z(G). Similarly, HP = G \leq C_G(H), so that H \leq Z(G). Thus HP = G \leq Z(G), and G is abelian.

(\Leftarrow) Suppose every group of order n = p_1^{a_1} \cdots p_k^{a_k} is abelian. If (without loss of generality) a_1 \geq 3, let Q be a nonabelian group of order p_1^{a_1}. (We may take Q to be the direct product of a nonabelian group of order p^3 and a cyclic group.) Then Q \times Z_{p_2^{a_2} \cdots p_k^{a_k}} is nonabelian of order n. Thus a_i \in \{1,2\} for all i. Now suppose (again without loss of generality) that p_1 divides p_2^{a_2} - 1. If a_2 = 1, then we may construct a nonabelian group Q of order p_1p_2. Then Q \times Z_{p_3^{a_3} \cdots p_k^{a_k}} is a nonabelian group of order n. If a_2 = 2, we may construct a nonabelian group of order pq^2, following the patterns here or here (depending on whether p_1 does not or does divide p_2-1.) By taking the direct product with a cyclic group, there exists a nonabelian group of order n.

Thus, if every group of order n is abelian, then n has the properties enumerated in the problem statement.

Basic properties of wreath products

Let K and L be groups, let n be a positive integer, let \rho : K \rightarrow S_n be a homomorphism, and let H = L^n. In this previous exercise, we defined an injective homomorphism \psi : S_n \rightarrow \mathsf{Aut}(H) by permuting the factors. Now \psi \circ \rho is a homomorphism K \rightarrow \mathsf{Aut}(H). The wreath product of L by K via \rho is defined L \wr_\rho K = H \rtimes_{\psi \circ \rho} K. Note that the wreath product depends on the choice of permutation representation; if none is given, \rho is assumed to be the left regular representation.

  1. Assume L and K are finite and \rho is the left regular representation of K. Find |L \wr_\rho K| in terms of |K| and |L|.
  2. Let p be a prime, let K = L = Z_p, and let \rho be the left regular representation of K. Prove that Z_p \wr_\rho Z_p is a nonabelian group of order p^{p+1} and is isomorphic to a Sylow p-subgroup of S_{p^2}. [The p copies of Z_p whose direct product makes up H may be represented by p disjoint p-cycles; these are cyclically permuted by K.]

  1. We have L \wr_\rho K = L^{|K|} \rtimes K. Thus |L \wr_\rho K| = |L|^{|K|} \cdot |K|.
  2. Since the left regular representation is nontrivial, and \psi is nontrivial, Z_p \wr_\rho Z_p is a nonabelian group. By part (1), this group has order p^p \cdot p = p^{p+1}.

    Now H = Z_p^p is a finite abelian group. Let \sigma_i, 1 \leq i \leq p, be p disjoint p-cycles in S_{p^2}. By this previous exercise, there is a unique group homomorphism \varphi : H \rightarrow S_{p^2} such that \varphi(e_i) = \sigma_i, where e_i is the ith “standard basis” vector. Moreover, (\psi \circ \rho)(k)(e_i) = e_{i+k}. By our solution to this previous exercise, there exists a permutation \tau \in S_{p^2} such that \tau \sigma_i \tau^{-1} = \sigma_{i+1}, where indices are taken mod p. Define \theta : Z_p \wr_\rho Z_p \rightarrow S_{p^2} by \theta(h,k) = \varphi(h)\tau^{-k}.

    I claim that \theta is a homomorphism. To see this, write h_1 = \prod e_i^{a_i} and h_2 = \prod e_i^{b_i}. Then \theta((h_1,k_1)(h_2,k_2)) = \theta(((\prod e_i^{a_i})(\psi \circ \rho)(k_1)(\prod e_i^{b_i}), k_1k_2)) = \theta(((\prod e_i^{a_i})(\prod e_{i+k_1}^{b_i}), k_1k_2)) = \theta((\prod e_i^{a_i + b_i - k_1}), k_1 k_2) = \prod \sigma_i^{a_i + b_i - k_1} \tau^{-k_1-k_2} = (\prod \sigma_i^{a_i})\tau^{-k_1} (\prod \sigma_i^{b_i}) \tau^{-k_2} = \theta(h_1,k_1) \theta(h_2,k_2).

    Moreover, I claim that \theta is injective; in this previous exercise we saw that every element of \mathsf{im}\ \theta can be written uniquely as a product (\prod \sigma_i^{a_i}) \tau^k. Thus \theta is injective. Counting elements, we see that \mathsf{im}\ \theta is a Sylow p-subgroup of S_{p^2}.

The group of upper triangular matrices over a field is the semidirect product of the diagonal matrices by upper triangular matrices of determinant 1

Let F be a field, let n be a positive integer, and let G = UT_n(F) be the group of upper triangular n \times n matrices over F.

  1. Prove that G = U \rtimes D, where U = \overline{UT}_n(F) is the group of strictly upper triangular matrices and D is the set of diagonal matrices in GL_n(F).
  2. Let n = 2. Recall that U \cong F and D \cong F^\times \times F^\times. Describe the homomorphism D \rightarrow \mathsf{Aut}(U) explicitly in terms of these isomorphisms. That is, show how each element of F^\times \times F^\times acts as an automorphism of F.

  1. Recall that SL_n(F) \leq GL_n(F) is normal; thus U = G \cap SL_n(F) is normal. Now let M = [m_{i,j}] \in G, and define A \in U and B \in D as follows: A = [a_{i,j}], where a_{i,j} = 1 if i=j, m_{i,j}/m_{j,j} if ij, then for all $latex k j$, b_{k,j} = 0. Thus c_{i,j} = 0. If i = j, then for all $latex k j$, b_{k,j} = 0. Thus c_{i,j} = m_{i,j}. If i<j, then for all k<i, a_{i,k} = 0, for i \leq k  j, b_{k,j} = 0. Thus c_{i,j} = a_{i,j}b_{j,j} = (m_{i,j}/m_{j,j})m_{j,j} = m_{i,j}. Thus we have AB = M. By the recognition theorem for semidirect products, G = U \rtimes D.
  2. Let B = \begin{bmatrix} \alpha & 0 \\ 0 & \beta \end{bmatrix} \in D and A = \begin{bmatrix} 1 & x \\ 0 & 1 \end{bmatrix} \in U. D acts on U by conjugation, and we have BAB^{-1} = \begin{bmatrix} 1 & \alpha x \beta^{-1} \\ 0 & 1 \end{bmatrix}. Thus the homomorphism \varphi : F^\times \times F^\times \rightarrow F is given by \varphi(\alpha,\beta)(x) = \alpha x \beta^{-1}.

For odd primes p, a p-group whose every subgroup is normal is abelian

Let p be an odd prime and let P be a p-group. Prove that if every subgroup of P is normal then P is abelian. (Note that Q_8 is a nonabelian 2-group with this property, so the result fails for p=2.)


[Disclaimer: I looked at Alfonso Gracia-Saz’s notes when solving this problem.]

We proceed by induction on k such that |P| = p^k.

For the base case, if k = 1, then P is cyclic and thus abelian.

For the inductive step, suppose the conclusion holds for all P such that |P| = p^k, where k \geq 1. Let P be a group of order p^{k+1}. If P is cyclic, then P is abelian. If P is not cyclic, then by the previous exercise there exists a normal subgroup U \leq P with U \cong Z_p \times Z_p. Choose x,y \in U such that U = \langle x \rangle \times \langle y \rangle. By the hypothesis, \langle x \rangle, \langle y \rangle \leq P are normal. Now P/\langle x \rangle and P/\langle y \rangle are groups of order p^k, and by the Lattice Isomorphism Theorem and the induction hypothesis, every subgroup of P/\langle x \rangle and P/\langle y \rangle is normal; thus P/\langle x \rangle and P/\langle y \rangle are abelian. By this previous exercise, P/(\langle x \rangle \cap \langle y \rangle) \cong P is abelian.

With p an odd prime, every noncyclic p-group contains a normal direct product of two copies of Cyc(p)

Let p be an odd prime. Prove that if P is a noncyclic p-group then P contains a normal subgroup U with U \cong Z_p \times Z_p. Deduce that for odd primes p a p-group that contains a unique subgroup of order p is cyclic. (Note: for p = 2, it is a theorem that the generalized quaternion groups Q_{2^n} are the only noncyclic 2-groups which contain a unique subgroup of order 2.)


Let p be an odd prime and let P be a p-group.

We will proceed by induction on the breadth of P; that is, on the number k such that |P| = p^k.

For the base case, let k = 2 and let P be a noncyclic group of order p^2. Then P \cong Z_p \times Z_p by Corollary 4.9. Then U = P is a normal subgroup of P isomorphic to Z_p \times Z_p.

For the inductive step, suppose that for some k \geq 2, every noncyclic group of order p^k contains a normal subgroup isomorphic to Z_p \times Z_p. Let P be a noncyclic group of order p^{k+1}. Since P is a p-group, Z(P) is nontrivial. By Cauchy’s Theorem, there exists an element x \in Z(P) of order p, and Z = \langle x \rangle is normal in P. Now P/Z is a p-group of order p^k.

Suppose P/Z is cyclic. By the Third Isomorphism Theorem, we have (P/Z)/(Z(P)/Z) \cong P/Z(P). Since P/Z is cyclic, every quotient of P/Z is cyclic, so that P/Z(P) is cyclic. By this previous exercise, P is abelian. Now P is a finite abelian group of rank at least 2, since P is not cyclic. Let \varphi : P \rightarrow P be the p-power map. By this previous exercise, \mathsf{ker}\ \varphi is the elementary abelian p-group whose rank is the same as that of P; that is, \mathsf{ker}\ \varphi \cong Z_p^t where t \geq 2. Clearly \mathsf{ker}\ \varphi contains a subgroup U isomorphic to Z_p \times Z_p, and since U \leq P and P is abelian, U is normal in P.

Suppose P/Z is not cyclic. By the induction hypothesis, there is a normal subgroup \overline{H} \leq P/Z such that \overline{H} \cong Z_p \times Z_p. By the Fourth (Lattice) Isomorphism Theorem, there is a normal subgroup H \leq P such that \overline{H} = H/Z. In particular, H is a noncyclic group of order p^3.

Now let \varphi : H \rightarrow H be the p-power map. By this previous exercise, \varphi is a homomorphism and in fact \mathsf{im}\ \varphi \leq Z(H). Note that H is not cyclic since H/Z is not cyclic. Thus, also by this exercise, \mathsf{ker}\ \varphi has order p^2 or p^3. Moreover, note that if x \in \mathsf{ker}\ \varphi and \alpha \in \mathsf{Aut}(H), then x has order p and thus \alpha(x) has order p. In particular, \alpha(x) \in \mathsf{ker}\ \varphi. So \alpha[\mathsf{ker}\ \varphi] \subseteq \mathsf{ker}\ \varphi, and since H is finite, in fact \alpha[\mathsf{ker}\ \varphi] = \mathsf{ker}\ \varphi. Thus \mathsf{ker}\ \varphi is characteristic in H, hence normal in P.

If |\mathsf{ker}\ \varphi| = p^2, then \mathsf{ker}\ \varphi is (isomorphic to) the elementary abelian group of order p^2; thus U = \mathsf{ker}\ \varphi \cong Z_p \times Z_p is normal in P.

[With a hint from Alfonso Gracia-Saz’s notes] Suppose |\mathsf{ker}\ \varphi| = p^3; then in fact H = \mathsf{ker}\ \varphi. Thus H is the elementary abelian group of order p^3, and H/Z \cong Z_p \times Z_p. Let \mathcal{H} denote the set of all subgroups K/Z \leq H/Z of order p. Note that P/Z acts on \mathcal{H} by conjugation, and that |\mathcal{H}| = p+1. Now given K/Z \in \mathcal{H}, by the Orbit-Stabilizer Theorem, we have |P/Z \cdot K| = [P/Z : \mathsf{stab}_{P/Z}(K/Z)]; since P/Z is a p-group, each orbit thus has order a power of p. Since |\mathcal{H}| = p+1, there is thus one orbit of order p and one orbit of order 1. Let N/Z be in the orbit of order 1. That is, N/Z is normal in P/Z, and by the Fourth Isomorphism Theorem, N \leq P is normal. Finally, |N| = p^2, and N \leq H is elementary abelian. Thus U = N \cong Z_p \times Z_p has the desired properties.

Finally, suppose P is a p group which contains a unique subgroup of order p. If P is not cyclic, then by the above argument P has a subgroup isomorphic to Z_p \times Z_p, yielding a contradiction because this subgroup contains many subgroups isomorphic to Z_p. Thus P is cyclic.

Some basic properties of holomorphs

Let H be a group of order n, let K = \mathsf{Aut}(H), and form G = \mathsf{Hol}(H) = H \rtimes_1 K. Let G act by left multiplication on the left cosets of K in G and let \pi : G \rightarrow S_{G/K} be the induced permutation representation.

  1. Prove that the elements of H are coset representatives for the left cosets of K in G and that with this choice of representatives, \pi|_H is the left regular representation of H.
  2. Prove that \mathsf{im}\ \pi is the normalizer in S_{G/K} of \pi[H]. Deduce that under the left regular representation of a finite group H of order n, the noramlizer in S_n of the image of H is isomorphic to \mathsf{Hol}(H). [Hint: Show that |G| = |N_{S_n}(\pi[H])|] using §5.5 #1 and §5.5 #2.
  3. Deduce that the normalizer of the group generated by an n-cycle in S_n is isomorphic to \mathsf{Hol}(Z_n) and has order n \varphi(n), where \varphi denotes the Euler totient function.

  1. Recall that (h,k) = (h,1)(1,k). It suffices to show that if h_1,h_2 \in H such that h_1K = h_2K, then h_1 = h_2. To that end, if (h_1,1) = (h_2,k), then h_1 = h_2. There are n left cosets of K, and H maps to these injectively. Thus H is a set of coset representatives of G/K. Define \theta : S_{G/K} \rightarrow S_H by \theta(hK) = h, and let \rho : H \rightarrow S_H be the left regular representation of H. Then \rho(h)(a) = ha = (ha)K h(aK) = \pi(h)(aK) = (\pi(h) \circ \theta)(a). Hence \rho(h) = \pi(h) \circ \theta; thus \pi[H] is the left regular representation of H up to an isomorphism.
  2. First, let \pi(g) \in \pi[G]. Now \pi(g)\pi[H]\pi(g)^{-1} = \pi[gHg^{-1}] = \pi[H]. Thus \pi(g) \in N_{S_{G/K}}(\pi[H]).

    Now we prove a lemma.

    Lemma [Adapted from Wikipedia]: If \sigma \in N_{S_{G/K}}(\pi[H]) and \sigma(1) = 1, then \sigma \in \mathsf{Aut}(H). Proof: Since \sigma normalizes \pi[H], for every h \in H there exists k \in H such that \sigma \circ \pi(h) = \pi(k) \circ \sigma. Applying this mapping to 1, we see that \sigma(h) = k. Thus, if \sigma(1) = 1, then for all h \in H we have \sigma \circ \pi(h) = \pi(\sigma(h)) \circ \sigma. Now for all h,k \in H, we have \sigma \circ \pi(h) \circ \pi(k) = \pi(\sigma(h)) \circ \sigma \circ \pi(k) = \pi(\sigma(h)) \circ \pi(\sigma(k)) \circ \sigma = \pi(\sigma(h) \sigma(k)) \circ \sigma, which is equal (since \pi is a homomorphism) to \sigma \circ \pi(hk) = \pi(\sigma(h,k)) \circ \sigma. Since \sigma is invertible and \pi is injective, we have \sigma(hk) = \sigma(h) \sigma(k); hence \sigma is a homomorphism H \rightarrow H. Since \sigma is also bijective, \sigma \in \mathsf{Aut}(H). \square

    (\supseteq) Now let \sigma \in N_{S_{G/K}}(\pi[H]). By the argument above, \pi(\sigma(1)^{-1}) also normalizes \pi[H]. Since the normalizer is a subgroup, we have \pi(\sigma^{-1}(1)) \circ \sigma \in N_{S_{G/K}}(\pi[H]). Moreover, (\pi(\sigma^{-1}(1)) \circ \sigma)(1) = \pi(\sigma(1)^{-1})(\sigma(1)) = 1; by the lemma, \beta = \pi(\sigma^{-1}(1)) \circ \sigma is an automorphism of H. Finally, note that \pi(\beta)(h) = \beta(h), so that \beta = \pi(\beta). Thus we have \sigma = \pi(\sigma(1)^{-1})^{-1} \circ \beta = \pi(\sigma(1)) \circ \pi(\beta) = \pi(\sigma(1),\beta). Hence \sigma \in \pi[G].

    Since \pi is injective, we conclude that \mathsf{Hol}(H) \cong N_{S_H}(\mathsf{im}\ \rho).

  3. Let \tau be an n-cycle in S_n. Up to a renaming, \langle \tau \rangle is the image of Z_n in S_{Z_n} via the permutation representation \pi : \mathsf{Hol}(Z_n) \rightarrow S_n used above. Now N_{S_n}(\langle \tau \rangle) \cong \mathsf{Hol}(Z_n), and \mathsf{Hol}(Z_n) = Z_n \rtimes \mathsf{Aut}(Z_n). We know that |\mathsf{Aut}(Z_n)| = \varphi(n), where \varphi denotes the Euler totient. Thus |N_{S_n}(\langle \tau \rangle)| = n\varphi(n).