Posts Tagged ‘abelian subgroups’

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 2k \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}.

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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).