## Maximum order of abelian subgroups in a symmetric group

Posted: April 21, 2012 in Elementary Algebra; Problems & Solutions, Groups and Fields
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For any set $X,$ we denote by ${\rm{Sym}}(X)$ the group of bijective maps from $X$ to $X.$

Problem. Let $\alpha : \mathbb{N} \longrightarrow [1, \infty)$ be any map which satisfies the following two conditions: $\alpha(p) \geq p$ and $\alpha(p+q) \geq \alpha(p) \alpha(q)$ for all $p,q \in \mathbb{N}.$ Let $X$ be a set with $|X|=n.$ Prove that if $G$ is an abelian subgroup of ${\rm{Sym}}(X),$ then $|G| \leq \alpha(n).$

Solution.  The proof is by induction on $n.$ If $n=1,$ then $|G|=1 \leq \alpha(1).$ Let $X$ be a set with $|X|= n \geq 2$ and suppose that the claim is true for any set of size $< n.$ Let $G$ be an abelian subgroup of ${\rm{Sym}}(X).$ Clearly $gx=g(x), \ g \in G, x \in X,$ defines an action of $G$ on $X.$ Let $X_1, \ldots , X_k$ be the orbits corresponding to this action and consider two cases.

Case 1. $k=1$: Fix an element $x_1 \in X.$ Then $X=X_1=Gx_1.$ Suppose that $g_1x_1=x_1$ for some $g_1 \in G$ and let $x \in X.$ Then $x=gx_1$ for some $g \in G.$ Thus, since $G$ is abelian, we have

$x=gx_1=gg_1x_1=g_1gx_1=g_1x.$

Hence $g_1x=x$ for all $x \in X$ and thus $g_1=1.$ So the stabilizer of $x_1$ is trivial and therefore, by the orbit-stabilizer theorem, $|G|=|X|=n \leq \alpha(n).$

Case 2$k \geq 2$: Let $|X_i|=n_i, \ i=1,2, \ldots, k.$ Clearly $\sum_{i=1}^k n_i=n$ and, since $k \geq 2,$ we have $n_i < n$ for all $i.$ For every $g \in G$ and $1 \leq i \leq k$ let $g_i=g|_{X_i},$ the restriction of $g$ to $X_i,$ and put

$G_i=\{g_i: \ g \in G\}.$

Then $g_i \in {\rm{Sym}}(X_i)$ and $G_i$ is an abelian subgroup of ${\rm{Sym}}(X_i).$ Thus, by the induction hypothesis

$|G_i| \leq \alpha(n_i),$

for all $i.$ Now, define $\varphi : G \longrightarrow \bigoplus_{i=1}^k G_i$ by $\varphi(g)=(g_1, g_2, \ldots , g_k)$ for all $g \in G.$ It is obvious that $\varphi$ is one-to-one and so

$|G| \leq |\bigoplus_{i=1}^k G_i|=\prod_{i=1}^k |G_i| \leq \prod_{i=1}^k \alpha(n_i) \leq \alpha(\sum_{i=1}^k n_i)=\alpha(n). \ \Box$

Remark. The map $\alpha: \mathbb{N} \longrightarrow [1, \infty)$ defined by $\alpha(p)=3^{p/3},$ for all $p \in \mathbb{N},$ satisfies both conditions in the above Problem. So if $|X|=n$ and if $G$ is an abelian subgroup of ${\rm{Sym}}(X),$ then $|G| \leq 3^{n/3}.$

## A finite non-abelian group in which every proper subgroup is abelian is not simple

Posted: June 8, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: , ,

The symmetric group $S_3$ is an example of a finite non-abelian group in which every proper subgroup is abelian. This group is not simple because its Sylow 3-subgroup is normal. In this post, we’ll show that this is the case for any finite (non-abelian) group all of whose proper subgroups are abelian.

Notation. Let $G$ be a group and let $H$ be a subgroup of $G.$ We will denote by $N(H)$ the normalizer of $H$ in $G.$ We will also define $\mathcal{C}(H)$ to be the union of conjugates of $H,$ i.e. $\mathcal{C}(H)=\bigcup_{g \in G} gHg^{-1}.$

Lemma 1. Let $G$ be a finite group and let $H$ be a subgroup of $G$ with this property that $gHg^{-1} \cap H = \{1\},$ for all $g \notin H.$ Then
1) the intersection of two distinct conjugates of $H$ is the identity element;
2) $|\mathcal{C}(H)|=|G|-[G:H] + 1$ and thus $|G \setminus \mathcal{C}(H)|=[G:H]-1.$

Proof. There is nothing to prove if $H=\{1\}.$ So suppose that $H \neq \{1\}.$ Clearly $H=N(H)$ because if $g \in N(H) \setminus H,$ then $H=gHg^{-1} \cap H = \{1\},$ which is a contradiction. Thus the number of conjugates of $H$ is $[G:N(H)]=[G:H].$
1) Suppose that $g_iHg_i^{-1}, \ i=1,2,$ are two distinct conjugates of $H$ and $g \in g_1Hg_1^{-1} \cap g_2Hg_2^{-1}.$ Then $g=g_1h_1g_1^{-1}=g_2h_2g_2^{-1},$ for some $h_1,h_2 \in H$ and hence $h_2 \in g_2^{-1}g_1 H g_1^{-1}g_2 \cap H.$ So, by hypothesis, either $h_2=1$ or $g_2^{-1}g_1 \in H.$ If $g_2^{-1}g_1 \in H,$ then $g_1Hg_1^{-1}=g_2Hg_2^{-1},$ which is a contradiction. Thus $h_2=1$ and so $g=1.$
2) Since every conjugate of $H$ has $|H|$ elements and, by 1), two distinct conjugates of $H$ have only one element in common , we have

$|\mathcal{C}(H)|=[G:H](|H|-1)+1=|G|-[G:H]+1. \ \Box$

Lemma 2. Let $G$ be a group and let $H_1$ and $H_2$ be two abelian subgroups of $G.$ Let $H$ be the subgroup generated by $H_1$ and $H_2.$ Then $H_1 \cap H_2$ is a normal subgroup of $H.$

Proof. Since $H_1$ and $H_2$ are abelian, $H_1 \cap H_2$ is a normal subgroup of both $H_1$ and $H_2.$ Thus $H_1 \cap H_2$ is also a normal subgroup of $H$ because an element of $H$ is a finite product of elements of $H_1$ and $H_2. \ \Box$

Problem. Let $G$ be a finite non-abelian group in which every proper subgroup is abelian. Prove that $G$ is not simple.

Solution. Suppose, to the contrary, that $G$ is simple and let $H$ be a maximal subgroup of $G.$ Clearly $N(H)=H$ because $H \subseteq N(H)$ and $H$ is a maximal subgroup of $G.$ Let $g \notin H.$ Then $g \notin N(H)$ and so $gHg^{-1} \neq H.$ Hence the subgroup generated by $H$ and $gHg^{-1}$ is $G$ because it contains $H$ strictly. Therefore $gHg^{-1} \cap H$ is a normal subgroup of $G,$ by Lemma 2. Hence $gHg^{-1} \cap H= \{1\}$ and so, by Lemma 1

$|G \setminus \mathcal{C}(H)|=[G:H]-1. \ \ \ \ \ \ \ \ (*)$

In particular $\mathcal{C}(H) \neq G$ and so we can choose $a \notin \mathcal{C}(H).$ Let $K$ be a maximal subgroup of $G$ which contains $a.$ Then again, exactly as we proved for $H,$ we have $gKg^{-1} \cap K = \{1\}$ and so , by Lemma 1

$|\mathcal{C}(K)|=|G|-[G:K]+1. \ \ \ \ \ \ \ \ \ (**)$

Now let $g_1,g_2 \in G.$ Note that $g_2Hg_2^{-1}$ is a maximal subgroup of $G$ because $H$ is a maximal subgroup of $G.$ Since $a \in K$ and $a \notin \mathcal{C}(H),$ we have $g_1Kg_1^{-1} \neq g_2Hg_2^{-1}$ and thus the subgroup generated by $g_1Kg_1^{-1}$ and $g_2Hg_2^{-1}$ is $G.$ Therefore $g_1Kg_1^{-1} \cap g_2Hg_2^{-1}$ is a normal subgroup of $G,$ by Lemma 2. Hence $g_1Kg_1^{-1} \cap g_2Hg_2^{-1}=\{1\}$ and so $\mathcal{C}(H) \cap \mathcal{C}(K)=\{1\}.$ Thus we have proved that

$\mathcal{C}(K) \setminus \{1\} \subseteq G \setminus \mathcal{C}(H).$

It now follows from $(*)$ and $(**)$ and the above inclusion that

$|G| \leq [G:H]+[G:K]-1 \leq |G|/2 + |G|/2 - 1 =|G|-1,$

which is non-sense. This contradiction shows that $G$ is not simple. $\Box$

In fact, the result still holds for finite non-cyclic groups because a simple abelian group is cyclic (of prime order).
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