## 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 $P^\prime = Z(P)$.