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.