Tag Archives: center (group)

The center of a subgroup of index 2 in a 2-group with cyclic center has rank at most 2

Use the preceding exercise to prove that if P is a 2-group which has a cyclic center and M is a subgroup of index 2 in P, then Z(M) has rank at most 2.


Note that M \leq P is normal and that Z(M) \leq M is characteristic; thus Z(M) is normal in P. Moreover, letting Z(M)_2 = \{ z \in Z(M) \ |\ z^2 = 1 \}, we know that Z(M)_2 \leq Z(M) is characteristic, so that Z(M)_2 \leq P is normal. Moreover, Z(M)_2 is elementary abelian by construction. Note that, by §5.2 #7, Z(M) and Z(M)_2 have the same rank.

If w \in P, then conjugation by w is an automorphism of Z(M)_2. If we choose w \notin M, and let z be fixed by w– that is, wz = zw, then we have z \in C_P(M) (since z \in Z(M)) and z \in C_P(\langle w \rangle). Since P = M \langle w \rangle, in fact w \in Z(P). Thus the subgroup of Z(M)_2 which is fixed under conjugation by w is contained in Z(P), and thus has rank 1.

By this previous exercise, Z(M)_2 has rank at most 2, so that Z(M) also has rank at most 2.

Compute the Frattini subgroup and number of maximal subgroups of a nonabelian group of order p³

Prove that if p is a prime and P is a nonabelian group of order p^3 then \Phi(P) = Z(P) and P/\Phi(P) \cong Z_p^2. Deduce that P has p+1 maximal subgroups.


Let P be a nonabelian group of order p. We now compute the upper central series of P. We have Z_0(P) = 1 and Z_1(P) = Z(P) \cong Z_p by this previous exercise. Also by that exercise, we know that P/Z_1(P) \cong Z_p^2 is abelian. Thus Z_2(P) = P, so that the upper central series of P is 1 < Z(P) < P.

By Theorem 8, the lower central series of P has length 3: 1 < P^\prime < P. Also by Theorem 8, we have P^\prime \leq Z(P); since P^\prime is nontrivial and Z(P) has prime order, in fact P^\prime = Z(P) (Note: we also proved this in this previous exercise). Using Burnside’s Basis Theorem, we know that \Phi(P) = P^\prime P^p = Z(P) P^p, where P^p = \{ x^p \ |\ x \in P. In this previous exercise, we showed that the p-power map on a nonabelian group of order p^3 is a homomorphism to Z(P), so that in fact P^p \leq Z(P) and we have \Phi(P) = Z(P).

By Lagrange, P/\Phi(P) has order p^2 and is elementary abelian. Thus P/\Phi(P) \cong Z_p \times Z_p has rank 2. By this previous theorem, P contains p+1 maximal subgroups.

Some properties of nonabelian p groups of order p³

Prove that if p is a prime and P a nonabelian group of order p^3, then |Z(P)| = p and P/Z(P) \cong Z_p \times Z_p.


By Lagrange, there are 4 possibilities for |Z(P)|: 1, p, p^2, and p^3. However, we know that Z(P) is nontrivial, and if |Z(P)| = p^3, then P is abelian.

If |Z(P)| = p^2, then P/Z(P) \cong Z_p is cyclic, and we have P abelian, a contradiction.

If |Z(P)| = p, there are two possibilities for the isomorphism type of P/Z(P): Z_{p^2} and Z_p \times Z_p. In the first case, P is abelian. Thus P/Z(P) \cong Z_p \times Z_p.

If a group mod its center is nilpotent, then the group is nilpotent

Prove that if G/Z(G) is nilpotent, then G is nilpotent.


We proved, as a lemma to the previous exercise, that Z_t(G/Z(G)) = Z_{t+1}(G)/Z(G) for all t. Now suppose G/Z(G) is nilpotent of nilpotence class k. Then G/Z(G) = Z_k(G/Z(G)) = Z_{k+1}(G)/Z(G), and we have G = Z_{k+1}(G). Thus G is nilpotent of nilpotence class at most k+1.

Some basic properties of nilpotent groups

Prove the following for G an infinite nilpotent group.

  1. Let G be a nilpotent group. If H \leq G is a nontrivial normal subgroup, then H \cap Z(G) is nontrivial. In particular, every normal subgroup of prime order is in the center.
  2. Let G be a nilpotent group. If H < G is a proper subgroup, then H < N_G(H) is a proper subgroup.

First we prove some lemmas.

Lemma 1: Let G be a group, and let Z_k(G) denote the k-th term in the upper central series of G. Then Z_k(G/Z(G)) = Z_{k+1}(G)/Z(G). Proof: We proceed by induction on k. For the base case, if k = 0, we have Z_0(G/Z(G)) = 1 = Z(G)/Z(G) = Z_1(G)/Z(G). For the inductive step, suppose Z_k(G/Z(G)) = Z_{k+1}(G)/Z(G) for some k \geq 0. We have the following chain of equivalent statements.

x Z(G) \in Z_{k+1}(G/Z(G))
\Leftrightarrow (xZ(G))Z_k(G/Z(G)) \in Z((G/Z(G))/Z_k(G/Z(G)))
\Leftrightarrow For all yZ(G) \in G/Z(G), (xyZ(G))Z_k(G/Z(G)) = (yxZ(G))Z_k(G/Z(G))
\Leftrightarrow For all yZ(G) \in G/Z(G), [x,y]Z(G) \in Z_k(G/Z(G))
\Leftrightarrow For all y \in G, [x,y]Z(G) \in Z_{k+1}(G)/Z(G)
\Leftrightarrow For all y \in G, [x,y] \in Z_{k+1}(G)
\Leftrightarrow For all y \in G, xy Z_{k+1}(G) = yx Z_{k+1}(G)
\Leftrightarrow x Z_{k+1}(G) \in Z(G/Z_{k+1}(G))
\Leftrightarrow x Z_{k+1}(G) \in Z_{k+2}(G)/Z_{k+1}(G)
\Leftrightarrow x \in Z_{k+2}(G)
\Leftrightarrow xZ(G) \in Z_{k+2}(G)/Z(G)

Thus Z_{k+1}(G/Z(G)) = Z_{k+2}(G)/Z(G), and the conclusion holds for all k. \square

Lemma 2: Let G be a group. If G is nilpotent of class k+1, then G/Z(G) is nilpotent of class k. Proof: G/Z(G) is nilpotent. Using Lemma 1, we have Z_k(G/Z(G)) = Z_{k+1}(G)/Z(G) = G/Z(G), so the nilpotence class of G/Z(G) is at most k. Now for t < k, we have Z_t(G/Z(G)) = Z_{t+1}(G)/Z(G); since G has nilpotence class k+1, Z_{t+1}(G) \neq G, and Z_t(G/Z(G)) \neq G/Z(G). Thus the nilpotence class of G/Z(G) is precisely k. \square

We now move to the main result.

  1. We proceed by induction on the nilpotence class of G.

    For the base case, if G has nilpotence class k =1, then G is abelian. Now any nontrivial normal subgroup H \leq G satisfies H \cap Z(G) = H \neq 1.

    For the inductive step, suppose the conclusion holds for any nilpotent group of nilpotence class k. G be a nilpotent group of class k+1, and let H \leq G be a nontrivial normal subgroup. Suppose by way of contradiction that H \cap Z(G) = 1; now consider the internal direct product HZ(G) \leq G. We have HZ(G)/Z(G) \leq G/Z(G); moreover, since H and Z(G) are normal, HZ(G)/Z(G) \leq G/Z(G) is normal. By Lemma 2, G/Z(G) is nilpotent of nilpotence class k, so that, by the induction hypothesis, HZ(G)/Z(G) \cap Z(G/Z(G)) is nontrivial. Let xzZ(G) \in HZ(G)/Z(G) \cap Z(G/Z(G)), where x \in H and z \in Z(G). In fact, we have xZ(G) \in Z(G/Z(G)). Let g \in G be arbitrary. Since H is normal in G, g^{-1}hg \in H. Now g^{-1}hgZ(G) = hZ(G) since hZ(G) is central in G/Z(G), so that h^{-1}g^{-1}hg \in Z(G). Recall, however, that H \cap Z(G) = 1, so that h^{-1}g^{-1}hg = 1, and we have gh = hg. Thus h \in Z(G), a contradiction. So H \cap Z(G) \neq 1.

  2. We proceed again by induction on the nilpotence class of G.

    For the base case, if G has nilpotence class k = 1, then G is abelian. Thus N_G(H) = G for all subgroups H, and if H < G is proper, then H < N_G(H) is proper.

    For the inductive step, suppose every nilpotent group of nilpotence class k \geq 1 has the desired property. Let G be a nilpotent group of nilpotence class k+1, and let H < G be a proper subgroup. Now G/Z(G) is nilpotent of class k. Suppose Z(G) \not\leq H; then, since Z(G) \leq N_G(H), H \leq \langle H, Z(G) \rangle is proper, and H \leq N_G(H) is proper. If Z(G) \leq H, then H/Z(G) \leq G/Z(G) is proper. By the induction hypothesis, H/Z(G) \leq N_{G/Z(G)}(H/Z(G)) = N_G(H)/Z(G) is proper, so that H \leq N_G(H) is proper.

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

In the Heisenberg group over a field, the commutator subgroup and center coincide

Let H(F) denote the Heisenberg group over a field F (cf. this previous exercise). Find an explicit formula for the commutator [X,Y], where X,Y \in H(F), and show that the commutator subgroup of H(F) equals the center of H(F).


In §1.4 #11 we found an explicit formula for the inverse of each X \in H(F).

Let X,Y \in H(F) with

X = \begin{bmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix} and
Y = \begin{bmatrix} 1 & d & e \\ 0 & 1 & f \\ 0 & 0 & 1 \end{bmatrix}.

Evidently,

[X,Y] = X^{-1}Y^{-1}XY
= \begin{bmatrix} 1 & -a & ac-b \\ 0 & 1 & -c \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & -d & df-e \\ 0 & 1 & -f \\ 0 & 0 & 1 \end{bmatrix} XY
= \begin{bmatrix} 1 & -d-a & df - e + af + ac - b \\ 0 & 1 & -f-c \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix} Y
= \begin{bmatrix} 1 & -d & df + af - cd - e \\ 0 & 1 & -f \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & d & e \\ 0 & 1 & f \\ 0 & 0 & 1 \end{bmatrix}
[X,Y] = \begin{bmatrix} 1 & 0 & af-cd \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

In particular, note that [X,Y] \in Z(H(F)) for all X,Y \in H(F). (We computed Z(H(F)) in this previous exercise.) Thus H(F)^\prime \leq Z(H(F)).

Note that to show equality, it suffices to show that for all b \in F, there exist a,f,c,d \in F such that b = af-cd. In fact, a = 1, f = b, and c = 0 suffice.

Thus H(F)^\prime = Z(H(F)).

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 p-power map on a group whose order is the cube of an odd prime

Let p be an odd prime and P a group of order p^3. Prove that the p-th power map x \mapsto x^p is a homomorphism P \rightarrow P and that \mathsf{im}\ \varphi \leq Z(P). If P is not cyclic, show that the kernel of \varphi has order p^2 or p^3.

Is the squaring map a homomorphism in nonabelian groups of order 8? Where is the oddness of p needed in the above proof?


If P is abelian, then we saw earlier that the p-power map is a homomorphism. Moreover \varphi trivially lands in Z(P) = P. If P is not cyclic, then by FTFGAG, P is isomorphic to either Z_p^3 or Z_{p^2} \times Z_p. By this previous exercise, the kernel of \varphi is an elementary abelian group of order p^3 or p^2, respectively.

If P is nonabelian, then by this previous exercise, P^\prime = Z(P) \cong Z_p. Note that \varphi(xy) = (xy)^p x^py^p[y,x]^{p(p-1)/2} = \varphi(x) \varphi(y) ([y,x]^p)^{(p-1)/2} = \varphi(x) \varphi(y). Thus \varphi is a homomorphism. Now note that for all x,y \in P, [y,x^p] = y^{-1}x^{-p}yx^p = (y^{-1}x^{-1}y)^px^p = (y^{-1}x^{-1}yx)^p = [y,x]^p = 1, since P^\prime \cong Z_p. Thus y \varphi(x) = \varphi(x) y, and we have \mathsf{im}\ \varphi \leq Z(P). Finally, by the First Isomorphism Theorem, P/\mathsf{ker}\ \varphi \cong \mathsf{im}\ \varphi \leq Z(P). Thus |P/\mathsf{ker}\ \varphi| \in \{1,p\}, and by Lagrange we have |\mathsf{ker}\ \varphi| \in \{p^2,p^3\}.

In a nonabelian p-group of order p³, the commutator subgroup and center are equal

Prove that if p is a prime and P a nonabelian group of order p^3, then P^\prime = Z(P).


Since P is nonabelian, we have Z(P) \neq P. Moreover, if |Z(P)| = p^2, then P/Z(P) \cong Z_p is cyclic, so that P is abelian; thus |Z(P)| \neq p^2. Since every p-group has a nontrivial center, we have |Z(P)| = p by Lagrange.

Now |P/Z(P)| = p^2; we know that every group of order p^2 is abelian, thus we have P^\prime \leq Z(P). Since P is nonabelian, P^\prime \neq 1; thus |P^\prime| \geq p, and we have P^\prime = Z(P).