## 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.