Tag Archives: normal subgroup

A finite group whose order is the product of three distinct primes has a normal Sylow subgroup of largest order

Let G be a group of order pqr where p < q < r and p, q, and r are primes. Prove that a Sylow r-subgroup of G is normal.


Recall that some Sylow subgroup of G is normal. Let P, Q, and R denote Sylow p-, q-, and r-subgroups of G, respectively.

Suppose n_p(G) = 1. Note that n_r(G/P) \in \{1,q\}, and since q < r, n_r(G/P) = 1. Let (via the lattice isomorphism theorem) \overline{R} \leq G be the subgroup whose quotient is the unique Sylow r-subgroup in G/P; we have |\overline{R}| = pr, and P \leq \overline{R} is normal. Moreover, because p < r, \overline{R} has a unique Sylow r-subgroup R, and we have \overline{R} \cong P \times R. Now if R^\prime \leq G is a Sylow r-subgroup, then R^\prime P/P = \overline{R}/P, so that \overline{R} = R^\prime P. Since R^\prime \leq \overline{R} is Sylow, R^\prime = R. Thus n_r(G) = 1.

The same argument works if n_q(G) = 1.

Every group of order 36 has a normal Sylow subgroup

Prove that if G is a group of order 36, then G has either a normal Sylow 2-subgroup or a normal Sylow 3-subgroup.


Note that 36 = 2^2 \cdot 3^2, so that Sylow’s Theorem forces n_2(G) \in \{1,3,9\} and n_3(G) \in \{1,4\}.

Suppose n_3(G) \neq 1. Then n_3(G) = 4 \not\equiv 1 mod 9, so that by Lemma 13 and this previous exercise, there exist P_3,Q_3 \in \mathsf{Syl}_3(G) such that P_3 \cap Q_3 is nontrivial. Consider now C_G(P_3 \cap Q_3); since P_3 and Q_3 are abelian, \langle P_3,Q_3 \rangle centralizes P_3 \cap Q_3, and moreover we can see by Sylow’s Theorem that G = \langle P_3, Q_3 \rangle. Thus P_3 \cap Q_3 \leq Z(G), and in fact P_3 \cap Q_3 is the intersection of all Sylow 3-subgroups of G. Thus G contains 2 + 6 + 6 + 6 + 6 = 26 elements of 3-power order. Let P_2 \leq G be a Sylow 2-subgroup; P_2 contains three nonidentity elements, whose products with the nonidentity elements in P_3 \cap Q_3 yield 6 elements of order 6 or 12. There are 4 elements left, which necessarily comprise a unique Sylow 2-subgroup.

No simple groups of order 144, 525, 2025, or 3159 exist

Prove that there are no simple groups of order 144, 525, 2025, or 3159.


  1. Note that 144 = 2^4 \cdot 3^2. Let G be a simple group of order 144. By Sylow’s Theorem, we have n_2 \in \{1,3,9\} and n_3 \in \{1,4,16\}. Note that |G| does not divide 5! since the largest power of 2 which divides 5! is 2^3. Thus no proper subgroup of G has index less than or equal to 5; in particular, n_2(G) = 9 and n_3(G) = 16. Note that n_3 \not\equiv 1 mod 9; by Lemma 13, there exist P_3, Q_3 \in \mathsf{Syl}_3(G) such that P_3 \cap Q_3 is normal in P_3 and Q_3. In particular, 3^2 divides |N_G(P_3 \cap Q_3)|, and since N_G(P_3 \cap Q_3) contains more than one Sylow 3-subgroup (namely P_3 and Q_3) its order is divisible by 2. In particular, [G : N_G(P_3 \cap Q_3)] \in \{1,2,4,8\}. Since G is simple and has no proper subgroups of index less than 6, in fact [G : N_G(P_3 \cap Q_3)] = 8, and we have |N_G(P_3 \cap Q_3)| = 2 \cdot 3^2.

    However, note that this implies that n_3(N_G(P_3 \cap Q_3)) = 1, by Sylow’s Theorem, but that we know N_G(P_3 \cap Q_3) contains at least two Sylow 3-subgroups. Thus we have a contradiction.

  2. Note that 525 = 3 \cdot 5^2 \cdot 7. Let G be a simple group of order 525. By Sylow’s Theorem, we have n_3 \in \{1,7,25,175\}, n_5 \in \{1,21\}, and n_7 \in \{1,15\}. Note that |G| does not divide 9! since the highest power of 5 dividing 9! is 5^1. Thus G has no proper subgroups of index at most 9. Moreover, n_5(G) = 21 \not\equiv 1 mod 25, so that by a previous exercise and Lemma 13, there exist P_5, Q_5 \in \mathsf{Syl}_5(G) such that P_5 \cap Q_5 \neq 1 and N_G(P_5 \cap Q_5) is divisible by 5^2 and some other prime.

    If |N_G(P_5 \cap Q_5)| \in \{ 3 \cdot 5^2, 7 \cdot 5^2\}, then G has a proper subgroup of index less than 9, a contradiction. If |N_G(P_5 \cap Q_5)| = 3 \cdot 5^2 \cdot 7, then P_5 \cap Q_5 is normal in G, a contradiction.

  3. Note that 2025 = 3^4 \cdot 5^2. Let G be a simple group of order 2025. By Sylow’s Theorem, we have n_3(G) = 25 and n_5(G) = 81. Note that |G| does not divide 9!, since the highest power of 5 dividing 9! is 5^1.

    Now since n_3(G) = 25 \not\equiv 1 mod 9, by Lemma 13 and a previous exercise there exist P_3, Q_3 \in \mathsf{Syl}_3(G) such that |N_G(P_3 \cap Q_3)| is divisible by 3^4 and 5. But then N_G(P_3 \cap Q_3) is either all of G or has index 5; in either case we have a contradiction.

  4. Note that 3159 = 3^5 \cdot 13. Let G be a simple group of order 3159. By Sylow’s Theorem we have n_3(G) = 13 and n_{13}(G) = 27. Note that n_3(G) = 13 \not\equiv 1 mod 9; thus by Lemma 13 and a previous exercise, there exist P_3, Q_3 \in \mathsf{Syl}_3(G) such that |N_G(P_3 \cap Q_3)| is divisible by 3^5 and 13- thus N_G(P_3 \cap Q_3) = G, a contradiction.

Every minimal normal subgroup of a finite solvable group is elementary abelian

For any group G a minimal normal subgroup is a normal subgroup M of G such that the only normal subgroups of G which are contained in M are M and 1. Prove that every minimal normal subgroup of a finite solvable group is an elementary abelian p-group for some prime p.


Let G be a finite solvable group and let N \leq G be a minimal normal subgroup.

If N=1, then the conclusion is trivial. Suppose N \neq 1.

Since G is solvable, N is solvable. Since N \neq 1, N^\prime < N is proper; now N^\prime is characteristic in N, hence normal in G. Since N is minimal, we have N^\prime = 1; thus N is abelian.

Recall from this previous exercise that N_p = \{ x \in N \ |\ x^p = 1 \} is characteristic in N, hence normal in G, for all primes p. Thus N_p \in \{1,N\} for all primes p. By Cauchy, some element of N has prime order, so that N_p \neq 1 for some prime p; in particular N_p = N. Thus N is an elementary abelian p-group.

The Frattini subgroup of a finite group is nilpotent

Let G be a finite group. Prove that \Phi(G) is nilpotent.


We will use Frattini’s Argument to show that every Sylow subgroup is normal.

Let P \leq \Phi(G) be a Sylow subgroup. By Frattini’s Argument, G = \Phi(G) N_G(P). By a lemma to a previous theorem, N_G(P) cannot be proper, so that N_G(P) = G and we have P \leq G normal. Then P \leq \Phi(G) is normal; thus all of the Sylow subgroups of \Phi(G) are normal, and thus \Phi(G) is nilpotent.

The Frattini subgroup is inclusion-monotone on normal subgroups

When G is a finite group prove that if N \leq G is normal, then \Phi(N) \leq \Phi(G). Give an explicit example when this containment does not hold if N is not normal in G.


[With help from the GroupProps wiki.]

We begin with some lemmas.

Lemma 1: Let G be a group, and let A,B,C \leq G with A \leq C be subgroups. Then A(B \cap C) = AB \cap C. Proof: (\subseteq) Let x \in A(B \cap C). Then x = ad where d \in B and d \in C, so that x \in AB \cap AC = AB \cap C. (\supseteq) Let x \in AB \cap C. Then we have x = ab = c for some a \in A, b \in B, and c \in C. Then b = a^{-1}c \in C, so in fact b \in B \cap C. Thus x \in A(B \cap C). \square

Lemma 2: Let G be a finite group. If H < G is a proper subgroup, then \Phi(G)H < G is proper. Proof: \Phi(G) \leq G is characteristic, hence normal, in G, so that \Phi(G)H \leq G is a subgroup. Now H is contained in some maximal subgroup M \leq G since G is finite, and \Phi(G) \leq M by definition. Thus \Phi(G)H \leq M < G is proper. \square

Now to the main result.

Let N \leq G be normal and suppose by way of contradiction that \Phi(N) \not\leq M for some maximal subgroup M \leq G. Since \Phi(N) \leq N is characteristic and N \leq G is normal, \Phi(N) \leq G is normal. Thus \Phi(N)M \leq G is a subgroup, and in fact \Phi(N)M = G.

Now N = G \cap N = \Phi(N)M \cap N = \phi(N)(M \cap N). However, since \Phi(N) \not\leq M, then N \not\leq M, so that M \cap N \leq N is proper. But this implies (by Lemma 2) that N < N is proper, a contradiction. Thus no such subgroup M exists, and in fact \Phi(N) \leq M for all maximal subgroups M \leq G. Hence \Phi(N) \leq \Phi(G).

Now before we produce a counterexample, we prove another lemma.

Lemma 3: \Phi(S_5) = 1. Proof: \Phi(S_5) is a proper normal subgroup, so that \Phi(S_5) \in \{1, A_5\}. Now S_5 contains a subgroup H isomorphic to S_4; by Lagrange, we have that [S_5,H] = 5 is prime, so that H is maximal. Recall also that A_5 is maximal; thus |\Phi(S_5)| divides both |H| = 24 and |A_5| = 60; hence |\Phi(S_5)| divides 12, and we have \Phi(S_5) = 1. \square

Consider K = \langle (1\ 2\ 3\ 4) \rangle \leq S_5; this subgroup is cyclic of order 4, and thus has a unique maximal subgroup which is isomorphic to Z_2. Thus K \leq S_5 but \Phi(K) \not\leq \Phi(S_5).

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.

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.

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.