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).

About these ads

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s