Tag Archives: sylow subgroup

Prove that a given tensor product of ZZ/(pⁿ) by a finite abelian group A is isomorphic to the Sylow p-subgroup of A

Let A be a finite abelian group of order n and let p^k be the largest power of the prime p dividing n. Prove that \mathbb{Z}/(p^k) \otimes_\mathbb{Z} A is isomorphic to the Sylow p-subgroup of A.


We saw in this previous exercise that as a \mathbb{Z}-module, A is the internal direct sum of its Sylow subgroups; say A = \bigoplus_Q M_q, where Q is the (finite) set of primes dividing n. Now tensor products commute with finite direct sums, so that \mathbb{Z}/(p^k) \otimes_\mathbb{Z} A \cong_\mathbb{Z} \mathbb{Z}/(p^k) \otimes_\mathbb{Z} (\bigoplus_Q M_q) \cong_\mathbb{Z} \bigoplus_Q (\mathbb{Z}/(p^k) \otimes_\mathbb{Z} M_q).

We claim that if q \neq p, then \mathbb{Z}/(p^k) \otimes_\mathbb{Z} M_q = 0. To see this, let |M_q| = q^\ell. Since q and p are relatively prime in the principal ideal domain \mathbb{Z}, there exist integers a and b such that aq^\ell + bp^k = 1. Now let \overline{x} \otimes y be a simple tensor in \mathbb{Z}/(p^k) \otimes_\mathbb{Z} M_q. We have \overline{x} \otimes y = (\overline{x} \cdot 1) \otimes y = \overline{x} \cdot (aq^\ell + bp^k) \otimes y = \overline{x} \cdot aq^\ell \otimes y + \overline{x} \cdot bp^k \otimes y = \overline{x} \cdot a \otimes q^\ell \cdot y + \overline{ap^k} \cdot b \otimes y = 0 + 0 = 0. Certainly then every element in this tensor product is zero, so that \mathbb{Z}/(p^k) \otimes_\mathbb{Z} M_q = 0.

Next, we claim that \mathbb{Z}/(p^k) \otimes_\mathbb{Z} M_p \cong_\mathbb{Z} M_p. To see this, note first that every simple tensor (hence every element) in \mathbb{Z}/(p^k) \otimes_\mathbb{Z} M_p can be written in the form \overline{1} \otimes m, since for an arbitrary simple tensor we have \overline{a} \otimes b = (\overline{1} \cdot a) \otimes b = \overline{1} \otimes a \cdot b. Now define \varphi : \mathbb{Z}/(p^k) \times M_p \rightarrow M_p by \varphi(\overline{a},m) = a \cdot m. Note that if p^k divides a-b, then (a-b) \cdot m = 0$, so that a \cdot m = b \cdot m. In particular, \varphi is well-defined. Evidently, \varphi is \mathbb{Z}-balanced, and so induces a group homomorphism \Phi : \mathbb{Z}/(p^k) \otimes_\mathbb{Z} M_p \rightarrow M_p. \Phi is surjective since \otimes{1} \otimes m \mapsto m. Since M_p and \mathbb{Z}/(p^k) \otimes_\mathbb{Z} M_p are finite sets, \Phi is a bijection and thus a group isomorphism.

Thus we have \mathbb{Z}/(p^k) \otimes_\mathbb{Z} A \cong_\mathbb{Z} M_p.

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.

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.

A sufficient condition for Sylow intersections of bounded index

Prove that if n_p \not\equiv 1 mod p^k, then there are distinct Sylow p-subgroups P and Q in G such that P \cap Q has index at most p^{k-1} in both P and Q.


(We will follow the strategy of this previous exercise.)

Let P \leq G be a Sylow p-subgroup. Now P acts on \mathsf{Syl}_p(G) by conjugation; note that if Q \in \mathsf{Syl}_p(G) is distinct from P, then using the Orbit-Stabilizer theorem and Lemma 4.19, we have |P \cdot Q| = [P : N_P(Q)] = [P : P \cap N_G(Q)] = [P : Q \cap P]. Since the orbit containing P itself has order 1, then n_p(G) = 1 + \sum [P : Q_i \cap P], where Q_i ranges over a set of orbit representatives. If every [P : Q_i \cap P] is divisible by p^k, then n_p(G) \equiv 1 mod p^k, a contradiction. Thus some [P : Q_i \cap P] is a power of p and is at most p^{k-1}.

We can see then that Q_i \cap P has index at most p^{k-1} in both P and Q_i.

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.

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.

If any Sylow p-subgroup intersects all others trivially, then all Sylow p-subgroups intersect trivially

Let G be a finite group and p a prime. Suppose that for some P \in \mathsf{Syl}_p(G), we have P \cap R = 1 for all R \in \mathsf{Syl}_p(G) with R \neq P. Prove that P_1 \cap P_2 = 1 for all P_1,P_2 \in \mathsf{Syl}_p(G). Deduce that in this case, the number of nonidentity elements of p-power order in G is precisely (|P|-1)[G:N_G(P)].


Let P_1 and P_2 be distinct Sylow p-subgroups. Now by Sylow’s Theorem, P_1 is conjugate to P in G; say xP_1x^{-1} = P. Now if we also have xP_2x^{-1} = P, then P_1 = x^{-1}Px = P_2, a contradiction. Note then that x(P_1 \cap P_2)x^{-1} = xP_1x^{-1} \cap xP_2 x^{-1} = P \cap Q, where Q is some Sylow p-subgroup of G distinct from P. Thus x(P_1 \cap P_2)x^{-1} = 1, and we have P_1 \cap P_2 = 1.

Now every nonidentity element of p-power order is in some Sylow p-subgroup of G, an no such element is in more than one. The number of nonidentity elements in a Sylow p-subgroup is |P|-1 and the number of Sylow p-subgroups is n_p(G) = [G:N_G(P)], and the final conclusion follows.

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.

Classification of groups of order 28

Classify the finite groups of order 28.


We begin with a lemma.

Lemma 1: Let H and K be groups, \varphi, \psi : K \rightarrow \mathsf{Aut}(H) group homomorphisms, and \theta : K \rightarrow K an automorphism. If \psi = \varphi \circ \theta, then the mapping \Phi : H \rtimes_\varphi K \rightarrow H \rtimes_\psi K given by \Phi((h,k)) = (h,\theta^{-1}(k)) is an isomorphism. Proof: (Homomorphism) Let (h_1,k_1), (h_2,k_2) \in H \rtimes_\varphi K. Then \Phi((h_1,k_1)(h_2,k_2)) = \Phi((h_1 \varphi(k_1)(h_2), k_1k_2)) = (h_1 \varphi(k_1)(h_2), \theta^{-1}(k_1k_2)) = (h_1 \varphi(\theta(\theta^{-1}(k_1)))(h_2), \theta^{-1}(k_1) \theta^{-1}(k_2)) = (h_1 (\varphi \circ \theta)(\theta^{-1}(k_1))(h_2), \theta^{-1}(k_1) \theta^{-1}(k_2)) = (h_1 \psi(\theta^{-1}(k_1))(h_2), \theta^{-1}(k_1) \theta^{-1}(k_2)) = (h_1, \theta^{-1}(k_1))(h_2, \theta^{-1}(k_2)) \Phi((h_1,k_1)) \Phi((h_2,k_2)). (Injective) If \Phi((h_1,k_1)) = \Phi((h_2,k_2)), then (h_1,\theta^{-1}(k_1)) = (h_2,\theta^{-1}(k_2)), so that h_1 = h_2 and k_1 = k_2. (Surjective) If (h,k) \in H \rtimes_\psi K, then (h,k) = \Phi((h, \theta(k))). \square

Now to the main result.

Note that 28 = 2^2 \cdot 7.

By FTFGAG, there are two distinct abelian groups of order 28: Z_{28} and Z_{14} \times Z_2.

Suppose now that G is a nonabelian group of order 28. By Sylow’s Theorem, we have n_7 = 1; thus G has a unique (hence normal) Sylow 7-subgroup H \cong Z_7. Let K \leq G be a Sylow 2-subgroup. By Lagrange, we have H \cap K = 1; thus G = HK, and by the recognition theorem for semidirect products, G \cong H \rtimes_\varphi K for some \varphi : K \rightarrow \mathsf{Aut}(H). Evidently, classifying the nonabelian groups of order 28 is equivalent to classifying the distinct groups constructed in this way. To that end, let H = Z_7 = \langle y \rangle. Then \mathsf{Aut}(Z_7) \cong Z_6 = \langle \alpha \rangle, where \alpha(x) = x^2.

Suppose K = Z_4 = \langle x \rangle. Now every homomorphism \varphi : Z_4 \rightarrow Z_6 is determined by \varphi(x), and moreover, provided \varphi(x)^4 = 1, any map thus defined is indeed a homomorphism. Thus we have two choices for \varphi(x): 1 and \alpha^3. If \varphi(x) = 1, then \varphi is trivial, contradicting the nonabelianicity of G. Thus \varphi(x) = \alpha^3 is the only group homomorphism K \rightarrow \mathsf{Aut}(H) which gives rise to a nonabelian group Z_7 \rtimes_\varphi Z_4 of order 28.

Suppose now that K = Z_2^2 = \langle a \rangle \times \langle b \rangle. Every homomorphism \varphi : K \rightarrow \mathsf{Aut}(H) is determined uniquely by \varphi(a) and \varphi(b), and provided \varphi(a)^2 = \varphi(b)^2 = 1, every mapping thus defined is a homomorphism. Thus we have \varphi(a), \varphi(b) \in \{1, \alpha^3 \}, for a total of four choices.

If \varphi_1(a) = \varphi_1(b) = 1, then \varphi is trivial, contradicting the nonabelianicity of G.

If \varphi_2(a) = \alpha^3 and \varphi_2(b) = 1, then Z_7 \rtimes_{\varphi_2} Z_2^2 is indeed a nonabelian group of order 28.

If \varphi_3(a) = 1 and \varphi_3(b) = \alpha^3, then we have \varphi_3 = \varphi_2 \circ \theta, where \theta(a) = b and \theta(b) = a. By the Lemma, H \rtimes_{\varphi_3} K \cong H \rtimes_{\varphi_2} K. Likewise, if \varphi_4(a) = \alpha^3 and \varphi_4(b) = \alpha^3, we do not get an essentially new group by the lemma, via the map \theta(a) = a and \theta(b) = ab.

Thus there is a unique nonabelian group of order 28 whose Sylow 2-subgroups are not cyclic.

Thus, the distinct groups of order 28 are as follows. In all cases, Z_7 = y, Z_4 = \langle x \rangle, and Z_2^2 = \langle a \rangle \times \langle b \rangle.

  1. Z_{28}
  2. Z_{14} \times Z_2
  3. Z_7 \rtimes_\varphi Z_4 where \varphi(x)(y) = y^{-1}
  4. Z_7 \rtimes_\psi Z_2^2 where \psi(a)(y) = y^{-1} and \psi(b)(y) = y