## Tag Archives: normalizer

### No simple groups of order 3³·7·13·409 exist

Given the information on the Sylow numbers for a simple group of order $3^3 \cdot 7 \cdot 13 \cdot 409$ found in the previous exercise, prove that there is no such group. [Hint: Work in the permutation representation of degree 819.]

Recall from the previous exercise that $n_{409}(G) = 819$, so that $|N_G(P_{409}| = 3 \cdot 409$. Now $N_G(P_{409}) = N$ acts on $\mathsf{Syl}_{409}(G) = S$ by conjugation. Moreover, since $N$ has a unique Sylow 409-subgroup, this action has precisely one orbit of order 1, namely $\{P_{409} \}$. By the Orbit Stabilizer Theorem and Lagrange, the remaining orbits have order 3, 409, or $3 \cdot 409$. Since $3 \cdot 409 > 819$, no orbit has this order. If some orbit has order 409, then there are $819 - 409 - 1 = 409$ remaining elements of $S$, which are distributed among orbits of order three. However, $409 \equiv 1$ mod 3, so there must be another orbit of order 1 – a contradiction. If no orbit has order 409, then there are $819 - 1 = 818$ elements of $S$ in orbits of order 3; however, $818 \equiv 2$ mod 3, again a contradiction. Thus no simple group of order $3^3 \cdot 7 \cdot 13 \cdot 409$ exists.

### Normal subsets of a Sylow subgroup which are conjugate in the supergroup are conjugate in the Sylow normalizer

Let $A$ and $B$ be normal subsets of a Sylow $p$-subgroup $P \leq G$. Prove that if $A$ and $B$ are conjugate in $G$ then they are conjugate in $N_G(P)$.

[With generous hints from Wikipedia.]

Suppose $B = gAg^{-1}$, with $g \in G$. Now since $P \leq N_G(A)$, $gPg^{-1} \leq N_G(gAg^{-1}) = N_G(B)$.

Note that $P, gPg^{-1} \leq N_G(B)$ are Sylow, and thus are conjugate in $N_G(B)$. So for some $h \in N_G(B)$, we have $P = hgPg^{-1}h^{-1}$, so that $hg \in N_G(P)$. Moreover, $hgAg^{-1}h^{-1} = hBh^{-1} = B$. Thus $A$ and $B$ are conjugate in $N_G(P)$.

### The normalizer of a maximal Sylow intersection is not a p-subgroup

Suppose that over all pairs of Sylow $p$-subgroups, $P$ and $Q$ are chosen so that $|P \cap Q|$ is maximal. Prove that $N_G(P \cap Q)$ is not a $p$-group.

We proved this in the course of answering this previous exercise.

### The normalizer of a maximal Sylow p-intersection contains more than one Sylow p-subgroup

Let $G$ be a group with more than one Sylow $p$-subgroup. Over all pairs of distinct Sylow $p$-subgroups let $P$ and $Q$ be chosen so that $|P \cap Q|$ is maximal. Show that $N_G(P \cap Q)$ has more than one Sylow $p$-subgroup and that any two distinct Sylow $p$-subgroups of $N_G(P \cap Q)$ intersect in $P \cap Q$. (Thus $|N_G(P \cap Q)|$ is divisible by $p|P \cap Q|$ and by some prime other than $p$. Note that Sylow $p$-subgroups of $N_G(P \cap Q)$ need not by Sylow in $G$.)

Note that $P \cap Q < N_P(P \cap Q)$ is proper since $P$ is a $p$-group. Since $N_P(P \cap Q) \leq N_G(P \cap Q)$, we have that $p|P \cap Q|$ divides $|N_G(P \cap Q)|$.

Suppose now that $N_G(P \cap Q)$ is a $p$-group. Then $N_G(P \cap Q) \leq R$ for some Sylow $p$-group $R$. If $R = P$, then $N_G(P \cap Q) \leq N_P(P \cap Q)$, and hence $N_G(P \cap Q) = N_P(P \cap Q)$. However, since $P \cap Q < N_Q(P \cap Q) \leq N_G(P \cap Q) \leq P$ and $P \cap Q < N_Q(P \cap Q) \leq Q$, we have a contradiction since there are elements in $P$ and $Q$ but not in $P \cap Q$ (namely, $N_Q(P \cap Q) \setminus P \cap Q$). Thus $R \neq P$. But then we have $P \cap Q < N_P(P \cap Q)$ $\leq N_G(P \cap Q)$ $\leq R$ and $P \cap Q < N_P(P \cap Q) \leq P$, so that $P \cap Q < N_P(P \cap Q) \leq P \cap R$. This implies that $P \cap Q$ is not (cardinality) maximal among the pairwise intersections of Sylow $p$-subgroups. Thus $N_G(P \cap Q)$ is not a $p$-group, so that its order is divisible by some prime other than $p$.

Suppose now that $N_G(P \cap Q)$ contains a unique Sylow $p$-subgroup, say $R$. Then $N_P(P \cap Q) \leq R$. Moreover, $R \leq P^\prime$ for some Sylow $p$-sugbroup $P^\prime$ of $G$. Since $P \cap Q < N_P(P \cap Q) \leq P \cap P^\prime$ and $P \cap Q$ is order maximal, we in fact have $R \leq P^\prime = P$. Similarly, $R \leq Q$. But then $P \cap Q \leq R \leq P \cap Q$, so that $P \cap Q$ is Sylow in $N_G(P \cap Q)$, a contradiction.

Let $A, B \in \mathsf{Syl}_p(N_G(P \cap Q))$. Since $P \cap Q$ is a $p$-group, $P \cap Q$ is contained in some Sylow $p$-group $R \leq N_G(P \cap Q)$. By Sylow’s Theorem, $R$ is conjugate to $A$ and $B$ in $N_G(P \cap Q)$, say by $x$ and $y$, respectively. Since $P \cap Q$ is normal in $N_G(P \cap Q)$, $P \cap Q \leq A, B$. Thus $P \cap Q \leq A \cap B$. Now $A \leq A^\prime$ and $B \leq B^\prime$ for some Sylow $p$-subgroups $A^\prime$ and $B^\prime$ of $G$. Suppose $A^\prime = B^\prime$. Then, since $A$ and $B$ are distinct, the subgroup $\langle A. B \rangle$ as generated in $A^\prime$ is a $p$-group, normalizes $P\cap Q$, and properly contains (for instance) $A$, which is a contradiction since $A$ is Sylow in $N_G(P \cap Q)$. Thus we have $A^\prime \neq B^\prime$. Moreover, $A \cap B \leq A^\prime \cap B^\prime$, and since $|P \cap Q|$ is maximal among the intersections of distinct Sylow $p$-subgroups and $G$ is finite, $P \cap Q = A \cap B$.

### No simple groups of order 2205, 4125, 5103, 6545, or 6435 exist

Prove that there are no simple groups of order 2205, 4125, 5103, 6545, or 6435.

We begin with a lemma.

Lemma 1: Let $G$ be a finite simple group. If $|G|$ does not divide $k!$, then for no subgroup $H \leq G$ do we have $[G:H] \leq k$. Proof: Suppose to the contrary that some $H$ exists, with $[G:H] \leq k$. This affords a left permutation representation $G \rightarrow S_{G/H}$, and the kernel of this representation is contained in $H$. Since the kernel is also normal in $G$, the kernel is trivial and the representation is injective; thus $G \leq S_{G/H} \leq S_k$. But by Lagrange this means that $|G|$ divides $k!$, a contradiction. $\square$

1. Note that $2205 = 3^2 \cdot 5 \cdot 7^2$. Let $G$ be a simple group of order 2205. Note that $|G|$ does not divide $13!$ since the largest power of 7 dividing $13!$ is $7^1$; by the lemma, $G$ has no subgroups of index less than or equal to 13. By Sylow’s Theorem, we have $n_7(G) = 15$. Note that $15 \not\equiv 1$ mod 49, so that by Lemma 13, there exist $P_7, Q_7 \in \mathsf{Syl}_7(G)$ such that $P_7 \cap Q_7 \neq 1$. Then $P_7 \cap Q_7 = P_0 \cong Z_7$. Moreover, $P_0$ is normal in $P_7$ and $Q_7$ since it is maximal in both; thus $P_7, Q_7 \leq N_G(P_0)$. Note that both $P_7$ and $Q_7$ are Sylow 7-subgroups of $N_G(P_0)$, so that $|N_G(P_0)| = 7^2 \cdot k$ where $k \geq 8$ (using Sylow). By Lagrange and because $G$ is simple, we have $k \in \{9,15\}$. But if $k=9$, then $N_G(P_0)$ has index 5, and if $k=15$, then $N_G(P_0)$ has index 3, both cases yielding a contradiction.
2. Note that $4125 = 3 \cdot 5^3 \cdot 11$. Let $G$ be a simple group of order 4125. By Sylow’s Theorem, we have $n_5(G) \in \{1,11\}$. Note that $|G|$ does not divide $14!$, since the largest power of 5 that divides $14!$ is $5^2$. By the lemma, no subgroup of $G$ has index less than 15. However, we have $n_5 = 11$, so that $[G:N_G(P_5)] = 11$ for each Sylow 5-subgroup $P_5$; this is a contradiction.
3. Note that $5103 = 3^6 \cdot 7$. Let $G$ be a simple group of order 5103. By Sylow’s Theorem we have $n_3(G) \in \{1,7\}$. Note that $|G|$ does not divide $8!$, since the highest power of 3 which divides $8!$ is $3^2$. By the lemma, no subgroup of $G$ has index 7, a contradiction since $[G:N_G(P_3)] = 7$ for each Sylow 3-subgroup $P_3$.
4. Note that $6545 = 5 \cdot 7 \cdot 11 \cdot 17$. Let $G$ be a simple group of order 6545. By Sylow’s Theorem, $n_5(G) \in \{1,11\}$. Now $|G|$ does not divide $16!$, so that by the lemma $G$ contains no subgroups of index 11. This is a contradiction since $[G:N_G(P_5)] = 11$ for each Sylow 5-subgroup $P_5$.
5. Note that $6435 = 3^2 \cdot 5 \cdot 11 \cdot 13$. Let $G$ be a simple group of order 6435. By Sylow’s Theorem, we have $n_5(G) \in \{1,11\}$. Now $|G|$ does not divide $12!$, so that by the lemma no subgroup has index 11. This is a contradiction since $[G:N_G(P_5)] = 11$ for each Sylow 5-subgroup $P_5$.

### Two consequences of Frattini’s argument regarding normalizers and centralizers

Let $G$ be a finite group, let $p$ be a prime, let $P \leq G$ be a Sylow $p$-subgroup, and let $N \leq G$ be a normal subgroup such that $p \not| |N|$. Prove the following.

1. $N_{G/N}(PN/N) = N_G(P)N/N$
2. $C_{G/N}(PN/N) = C_G(P)N/N$

Note that $P \cap N = 1$, so that $PN \cong P \times N$. In particular, $P \leq C_G(N)$ and $N \leq C_G(P)$.

Note that $P \leq PN \cong P \times N$ is a Sylow subgroup since $p$ does not divide $|N|$, and that $PN \leq N_G(PN)$ is normal. By Frattini’s Argument, we have $N_G(PN) = PN N_G(P) = N_G(P)N$. Thus $N_G(P)N/N = N_G(PN)/N = N_{G/N}(PN/N)$, as desired.

Now note that $C_G(PN) \leq C_G(P)$, so that $C_{G/N}(PN/N) = C_G(PN)/N \leq C_G(P)N/N$. Moreover, we have $[C_G(P)/N, PN/N] = [C_G(P),PN]/N$ $= [C_G(P),N]/N$ $= 1$. Thus $C_G(P)/N \leq C_{G/N}(PN/N)$.

### 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. $\square$

### The centralizer and normalizer of a nonnormal semidirect factor in a normal semidirect factor are equal

Let $H$ and $K$ be groups, let $\varphi : K \rightarrow \mathsf{Aut}(H)$ be a group homomorphism, and identify $H$ and $K$ as subgroups of $G = H \rtimes_\varphi K$.

Prove that $C_H(K) = N_H(K)$.

Recall that $C_H(K) \leq N_H(K)$ always holds.

$(\supseteq)$ Let $(h,1) \in N_H(K)$. Then for every $(1,k) \in K$, $(h,1)^{-1}(1,k)(h,1) \in K$. Carrying out this multiplication, we see that $(h^{-1} \varphi(k)(h),k) \in K$. Thus $h^{-1}\varphi(k)(h) = 1$, and we have $\varphi(k)(h) = h$. Then $(\varphi(k)(h),k) = (h,k)$, and evidently $(1,k) \cdot (h,1) = (h,1) \cdot (1,k)$. Thus $(h,1) \in C_H(K)$.

### A congruence condition on the index of subgroups containing Sylow normalizers

Let $P$ be a Sylow $p$-subgroup of $G$ and let $M$ be any subgroup of $G$ which contains $N_G(P)$. Prove that $[G : M] \equiv 1 \mod p$.

We prove a slightly stronger result by induction. First a definition.

Let $G$ be a finite group and $H \leq G$ a subgroup. The distance from $H$ to $G$ is the length $d$ of the longest chain of proper subgroups $K_i$ such that $H = K_0 < K_1 < \ldots < K_d$.

Lemma: Let $G$ be a finite group and $P \leq G$ a Sylow $p$-subgroup. If $H,K \leq G$ are subgroups such that $N_G(P) \leq H \leq K$, then $[K : H] \equiv 1 \mod p$. Proof: We proceed by induction on the distance $d$ from $N_G(P)$ to $K$. For the base case, if $d = 0$, then $K = N_G(P)$ so that $H = N_G(P)$, and we have $[K : H] = 1$. For the inductive step, suppose that for some $d \geq 0$, if the distance from $N_G(P)$ to $K$ is at most $d$ then for all $H$ such that $N_G(P) \leq H \leq K$ we have $[K : H] \equiv 1 \mod p$. Now let $K \leq G$ be a subgroup containing $N_G(P)$ such that the distance from $N_G(P)$ to $K$ is $d+1$, and suppose further that $N_G(P) \leq H \leq K$. First, if $H = K$, we have $[K : H] = 1$ so that the conclusion holds. Suppose now that the inclusion $H \leq K$ is proper. Note that $[K : N_G(P)] = [K : H] \cdot [H : N_G(P)]$. By the induction hypothesis, we have $[H : N_G(P)] \equiv 1 \mod p$. Recall that $N_K(P) = K \cap N_G(P)$, so that $[K : N_G(P)] = [K : N_K(P)]$ $= n_p(K) \equiv 1 \mod p$ by Sylow's Theorem. Modulo $p$, we thus have $[K:H] \equiv 1$. $\square$

In particular, if $K = G$ and $N_G(P) \leq M \leq G$, we have $[G:M] \equiv 1 \mod p$.