Tag Archives: quotient group

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).

If the first two lower central quotients of a group are cyclic, then the second derived subgroup is trivial

Show that if G^\prime/G^{\prime\prime} and G^{\prime\prime} / G^{\prime\prime\prime} are both cyclic then G^{\prime\prime} = 1.

We begin by proving a lemma.

Lemma: Let G be a group and let N \leq G be normal. Then (G/N)^\prime = G^\prime N /N. Proof: (\subseteq) If [xN,yN] \in G/N is a commutator, then [xN,yN] = [x,y]N \in G^\prime N/N. (\supseteq) G^\prime \leq G^\prime N, so that G/G^\prime N \cong (G/N)/(G^\prime N/N) is abelian. By Theorem 5.7 in the text, (G/N)^\prime \leq G^\prime N/N. \square

Now for the main result.

Note that G^{\prime\prime\prime} \leq G is characteristic, hence normal, so that G^{\prime\prime}/G^{\prime\prime\prime} \leq G/G^{\prime\prime\prime} is a cyclic, hence abelian, normal subgroup.

By this previous exercise, (G/G^{\prime\prime\prime})/(G^{\prime\prime}/G^{\prime\prime\prime}) \cong G/G^{\prime\prime} (via the Third Isomorphism Theorem) acts on G^{\prime\prime}/G^{\prime\prime\prime} by conjugation on the left as follows: (gG^{\prime\prime}) \cdot (a G^{\prime\prime\prime}) = (gag^{-1}) G^{\prime\prime\prime}. Now conjugation is an automorphism of G^{\prime\prime}/G^{\prime\prime\prime}, and since this quotient is cyclic, its automorphism group is abelian. Thus, for all g,h \in G and a \in G^{\prime\prime}, we have (gh)a(gh)^{-1} G^{\prime\prime\prime} = (hg)a(hg)^{-1} G^{\prime\prime\prime}. Via some arithmetic, we see that [g,h] a [g,h]^{-1} a^{-1} \in G^{\prime\prime\prime}; in particular, each element of G^{\prime\prime}/G^{\prime\prime\prime} commutes with each generator of G^{\prime}/G^{\prime\prime\prime}, so that G^{\prime\prime}/G^{\prime\prime\prime} \leq Z(G^{\prime}/G^{\prime\prime\prime}).

Using the Third Isomorphism Theorem, (G^\prime/G^{\prime\prime\prime})/Z(G^\prime/G^{\prime\prime\prime}) \cong ((G^\prime/G^{\prime\prime\prime})/(G^{\prime\prime}/G^{\prime\prime\prime}))/(Z(G^\prime/G^{\prime\prime\prime})/(G^{\prime\prime}/G^{\prime\prime\prime})) \cong (G^\prime/G^{\prime\prime})/(Z(G^\prime/G^{\prime\prime\prime})/(G^{\prime\prime}/G^{\prime\prime\prime})) is cyclic; hence G^\prime/G^{\prime\prime\prime} is abelian, and using the lemma we have 1 = (G^\prime/G^{\prime\prime\prime})^\prime = G^{\prime\prime}/G^{\prime\prime\prime}. Thus G^{\prime\prime} = G^{\prime\prime\prime}.

The class of nilpotent groups is closed under subgroups and quotients, but not extensions

Prove that subgroups and quotient groups of nilpotent groups are nilpotent. (Your proof should work for infinite groups.) Give an explicit example of a group G which possesses a normal subgroup H such that H and G/H are nilpotent but G is not nilpotent.

We showed that subgroups and quotients of nilpotent groups are nilpotent in the lemmas to this previous exercise.

Consider now D_6. \langle r \rangle \leq D_6 is a normal subgroup of order 3, and thus \langle r \rangle \cong Z_3 is abelian, hence nilpotent. Moreover, D_6/\langle r \rangle \cong Z_2 is abelian, hence nilpotent. However, D_6 has distinct Sylow 2-subgroups \langle s \rangle and \langle sr \rangle (for example). By Theorem 3, D_6 is not nilpotent.

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.

If the quotients of a group by two subgroups are abelian, then the quotient by their intersection is abelian

Let A,B \leq G be normal subgroups such that G/A and G/B are abelian. Prove that G/(A \cap B) is abelian.

Since G/A is abelian we have [G,G] \leq A. Likewise, [G,G] \leq B. Thus [G,G] \leq A \cap B, so that G/(A \cap B) is abelian.

Compute a given quotient of a free abelian group

Let n and k be positive integers and let A be the free abelian group of rank n written additively. Prove that A/kA is isomorphic to (\mathbb{Z}/k\mathbb{Z})^n, where kA = \{ ka \ |\ a \in A \}.

We have A = \mathbb{Z}^n. Note that x \in k \mathbb{Z}^n, then x = k(a_i), so that x = (ka_i), hence x \in (k\mathbb{Z})^n.

By this previous exercise, \mathbb{Z}^n / k\mathbb{Z}^n \cong (\mathbb{Z}/(k))^n.

A quotient by a product is isomorphic to the product of quotients

Let G = A_1 \times A_2 \times \cdots \times A_n and for each i let B_i \leq A_i be a normal subgroup. Prove that B_1 \times \cdots \times B_n \leq G is normal and that (A_1 \times \cdots \times A_n)/(B_1 \times \cdots \times B_n) \cong (A_1/B_1) \times \cdots \times (A_n/B_n).

We begin with some lemmas.

Lemma 1: Let \varphi_1 : G_1 \rightarrow H_1 and \varphi_2 : G_2 \rightarrow H_2 be group homomorphisms. Then \mathsf{ker}(\varphi_1 \times \varphi_2) = (\mathsf{ker}\ \varphi_1) \times (\mathsf{ker}\ \varphi_2). Proof: (\subseteq) Let (a,b) \in \mathsf{ker}(\varphi_1 \times \varphi_2). Then (1,1) = (\varphi_1 \times \varphi_2)(a,b) = (\varphi_1(a), \varphi_2(b)), hence \varphi_1(a) = 1 and \varphi_2(b) = 1. Thus a \in \mathsf{ker}\ \varphi_1 and b \in \mathsf{ker}\ \varphi_2, and (a,b) \in (\mathsf{ker}\ \varphi_1) \times (\mathsf{ker}\ \varphi_2). (\supseteq) If (a,b) \in (\mathsf{ker}\ \varphi_1) \times (\mathsf{ker}\ \varphi_2), then (\varphi_1 \times \varphi_2)(a,b) = (1,1); hence (a,b) \in \mathsf{ker}(\varphi_1 \times \varphi_2). \square

Lemma 2: Let A_1 and A_2 be groups with normal subgroups B_1 \leq A_1 and B_2 \leq A_2. Then B_1 \times B_2 is normal in A_1 \times A_2 and (A_1 \times A_2)/(B_1 \times B_2) \cong (A_1/B_1) \times (A_2/B_2). Proof: Let \pi_1 : A_1 \rightarrow A_1/B_1 and \pi_2 : A_2 \rightarrow A_2/B_2 denote the natural projections. These mappings are surjective, so that \pi_1 \times \pi_2 is a surjective homomorphism A_1 \times A_2 \rightarrow (A_1/B_1) \times (A_2/B_2). By Lemma 1, we have \mathsf{ker}(\pi_1 \times \pi_2) = B_1 \times B_2. The conclusion follows by the First Isomorphism Theorem. \square

The main result now follows by induction on n.

An upper bound on the number of Sylow subgroups of a finite quotient

Let G be a finite group and suppose N \leq G is normal. Prove that n_p(G/N) \leq n_p(G).

We begin with a lemma.

Lemma: Every Sylow p-subgroup of G/N has the form PN/N for some Sylow p-subgroup P of G. Proof: Let P be a Sylow p-subgroup of G. In this previous exercise, we saw that PN/N is a Sylow p-subgroup of G/N. Now let Q \leq G/N be any Sylow p-subgroup; then Q = (gN)(PN/N)(g^{-1}N) = (gPg^{-1})N/N for some g \in G by Sylow’s Theorem, and gPg^{-1} is a Sylow p-subgroup of G. \square

Now to the main result.

Note that the mapping \Phi : \mathsf{Syl}_p(G) \rightarrow \mathsf{Syl}_p(G/N) given by P \mapsto PN/N is a surjection of finite sets. Thus n_p(G) \geq n_p(G/N).

In a finite group, Sylow subgroups induce Sylow subgroups in normal subgroups and their corresponding quotients

Let G be a finite group, P \in \mathsf{Syl}_p(G), and N \leq G a normal subgroup. Use the conjugacy part of Sylow’s Theorem to prove that P \cap N is a Sylow p-subgroup of N. Deduce that PN/N is a Sylow p-subgroup of G/N.

Let Q be a Sylow p-subgroup of N. Q is a p-subgroup of G, so that Q \leq xPx^{-1} for some x \in G. Then Q \leq xPx^{-1} \cap N = xPx^{-1} \cap xNx^{-1} since N is normal, hence Q \leq x(P \cap N)x^{-1}, thus x^{-1}Qx \leq P \cap N. Now x^{-1}Qx is a subgroup of N, and has maximal p-power order in N, since |x^{-1}Qx| = |Q|. Since P \cap N is a p-subgroup of N, P \cap N is a maximal p-power subgroup of N, thus a Sylow p-subgroup.

Note that PN/N \cong P/(P \cap N) by the Second Isomorphism Theorem, so that PN/N is a p-subgroup of G/N. By the Third Isomorphism Theorem, [G/N : PN/N] = [G : PN]; since P is a maximal p-power order subgroup of G, p does not divide [G : PN], so p does not divide [G/N : PN/N]. Hence PN/N is a Sylow p-subgroup of G/N.

An action of a quotient group on an abelian normal subgroup

Let G be a group and let A be an abelian, normal subgroup of G. Show that G/A acts (on the left) by conjugation on A by [g] \cdot a = gag^{-1}. (In particular, show that this action is well defined.) Give an explicit example to show that this action is not well defined if A is nonabelian.

To see well definedness, suppose gh^{-1} = b \in A. That is, g = bh. Now gag^{-1} = bhah^{-1}b^{-1}; note that hah^{-1} \in A since A is normal, hence gag^{-1} = hah^{-1} since A is abelian. Thus [g] \cdot a = [h] \cdot a, and the action is well defined.

Clearly we have [1] \cdot a = a and [gh] \cdot a = ghah^{-1}g^{-1} = [g] \cdot ([h] \cdot a) = ([g][h]) \cdot a. So G/A acts on A.

Now identify D_6 = \langle r^2, s \rangle in D_{12}. Note that D_6 \leq D_{12} has index 2, hence is normal, and that D_6 is nonabelian. Moreover, r and rs are distinct representatives of rD_6. However, rr^2r^{-1} = r^2 and rsr^2rs = r^4, so that the action described above is not well defined.