Characteristically simple groups

Posted: June 12, 2024 in Basic Algebra, Groups
Tags: characteristic subgroup, characteristically simple groups, direct product of groups, elementary abelian p-group, minimal normal subgroup, non-abelian simple group, solvable group

Let $G$ be a group. A subgroup $H$ of $G$ is called characteristic if $f(H)=H$ for every automorphism of $f$ of $G.$ Compare that to the definition of a normal subgroup: a subgroup $H$ of $G$ is normal if $f(H)=H$ for every inner automorphism $f$ of $G.$ So every characteristic subgroup is normal. A few examples of characteristic subgroups of $G$ are: the trivial subgroup $(1),$ the group $G$ itself, the center of $G,$ and the commutator subgroup of $G.$

Remark. Let $H$ be a subgroup of a group $G.$ If $f$ is an automorphism of $G,$ then $f^{-1}$ is also an automorphism of $G.$ So $H$ is characteristic if and only if $f(H) \subseteq H$ for all automorphisms $f$ of $G.$

Definition 1. A group $G \ne (1)$ is called characteristically simple if the only characteristic subgroups of $G$ are $(1)$ and $G$ itself.

In this post, we characterize all finite characteristically simple groups. But first, let’s give a few examples of those groups. The trivial ones are simple groups; it is clear that every simple group is characteristically simple because every characteristic subgroup is normal. A highly non-trivial example is given in Example 2 in this post. A slightly non-trivial examples of characteristically simple groups are finite elementary abelian $p$ -groups. Recall that a finite elementary abelian $p$ –group, where $p$ is a prime number, is any finite direct product of copies of the cyclic group $\mathbb{Z}_p.$ Looking at the group additively, it is clear that finite elementary abelian $p$ -groups are just finite dimensional vector spaces over the finite field $\mathbb{Z}_p$ (or you can call this one $\mathbb{F}_p$ if you have trouble distinguishing between $\mathbb{Z}_p$ as a cyclic group and $\mathbb{Z}_p$ as a field).

Example 1. Every finite elementary abelian $p$ -group $G$ is characteristically simple.

Proof. So, considering $G$ additively, $G$ is a finite dimensional vector space over the field $\mathbb{Z}_p.$ Suppose, to the contrary, that $G$ has a characteristic subgroup $H \ne (1), G.$ Let $x_1, \cdots , x_m$ be a basis for $H$ and extend it to a basis $x_1, \cdots , x_m, x_{m+1}, \cdots , x_n$ of $G.$ Now, the $\mathbb{Z}_p$ -linear map $f: G \to G$ defined by

$f(x_1)=x_n, \ f(x_n)=x_1, \ f(x_i)=x_i, \ i \ne 1,n,$

is an automorphism of $G$ and so, since $H$ is characteristic, $x_n=f(x_1) \in f(H)=H,$ contradiction. $\ \Box$

Definition 2. A minimal normal subgroup of a group $G$ is a normal subgroup $H \ne (1)$ of $G$ such that the only subgroups of $H$ that are normal in $G$ are $(1)$ and $H$ itself.

Example 2. Every characteristic subgroup of a normal subgroup is normal. In particular, every minimal normal subgroup of a group is characteristically simple.

Proof. Let $K$ be a characteristic subgroup of a normal subgroup $H$ of a group $G,$ and let $g \in G.$ Then, since $H$ is a normal subgroup of $G,$ the map $f: H \to H$ defined by $f(x)=gxg^{-1}$ is an automorphism of $H.$ Since $K$ is a characteristic subgroup of $H,$ we have $f(K)=K$ and so $gxg^{-1} \in K$ for all $x \in K,$ i.e. $K$ is normal in $G. \ \Box$

We are now ready to characterize all finite characteristically simple groups.

Theorem. A finite group $G$ is characteristically simple if and only if $G$ is the direct product of some isomorphic simple groups.

Proof. Suppose first that

$G=\underbrace{S \times S \times \cdots \times S}_{n \ \text{times}},$

where $S$ is a simple group. We need to show that $G$ is characteristically simple. Since $S$ is simple, either $S$ is abelian or $Z(S),$ the center of $S,$ is trivial. If $S$ is abelian, then $S \cong \mathbb{Z}_p,$ for some prime $p,$ and hence $G$ is characteristically simple by Example 1. So we may assume that $Z(S)=(1).$ Suppose that $H \ne (1)$ is a characteristic subgroup of $G,$ and let $h= (s_1, \cdots , s_k, \cdots , s_n) \in H, \ s_i \in S, s_k \ne 1.$ Then $s_k \notin Z(S)$ and so there exists $t \in S$ such that $s_kt \ne ts_k.$ Let $g:=(1, \cdots , t, \cdots , 1) \in G.$ Then, since $H$ is normal,

$(1, \cdots , s_k^{-1}ts_kt^{-1}, \cdots ,1)=h^{-1}ghg^{-1} \in H.$

Notice that since $s_kt \ne ts_k,$ we have $s_k^{-1}ts_kt^{-1} \ne 1.$ So we have shown the set

$S_k:=\{x_k \in S: \ (1, \cdots , x_k, \cdots ,1) \in H\},$

which is a normal subgroup of $S,$ is non-trivial and hence $S_k=S,$ i.e., $(1, \cdots , x_k, \cdots , 1) \in H$ for all $x_k \in S.$ So $(1, \cdots , x_j, \cdots , 1) \in H$ for all $j$ and all $x_j \in S$ because the map $f: G \to G$ defined by

$f(x_1, \cdots , x_j, \cdots , x_k, \cdots , x_n)=(x_1, \cdots , x_k, \cdots , x_j, \cdots , x_n)$

is an automorphism of $G$ and $H$ is characteristic. Thus $H=G$ and the proof of one side of the Theorem is complete. Notice that for this part of the Theorem, we did not need $G$ to be finite.

Conversely, suppose that $G$ is characteristically simple. We need to show that $G$ is the direct product of some isomorphic simple groups. Choose a normal subgroup $H$ of $G$ such that $1 < |H| \le |G|$ and $|H|$ is as small as possible. Let $N, N_1, \cdots , N_k$ be subgroups of $G$ such that $N=N_1 \times \cdots \times N_k, \ N_i \cong H, \ \forall i,$ and the positive integer $k$ is as large as possible.

Claim 1. Let $f$ be an automorphism of $G.$ Then $f(N_i) \subseteq N$ for all $i$

Proof. First notice that since $N_i \cong H$ for all $i,$ each $N_i,$ hence $N,$ is a normal subgroup of $G$ and $|N_i|=|H|$ for all $i.$ Suppose now, to the contrary, that $f(N_i) \nsubseteq N$ for some $i.$ So $f(N_i) \cap N$ is a proper subgroup of $f(N_i)$ and thus $|f(N_i) \cap N| < |f(N_i)|=|N_i|=|H|.$ Hence, since $f(N_i) \cap N$ is a normal subgroup of $G,$ we have $f(N_i) \cap N=(1),$ by minimality of $|H|.$ So $f(N_i),N$ are two normal subgroups of $G$ whose intersection is trivial. That implies $Nf(N_i)=N \times f(N_i)=N_1 \times \cdots N_k \times f(N_i),$ which contradicts maximality of $k.$ This contradiction completes the proof of Claim 1.

Claim 2. $N=G.$

Proof. By Claim 1, $f(N_i) \subseteq N$ for all $i,$ and so $f(N) \subseteq N$ for all automorphisms $f$ of $G.$ Therefore $N$ is a characteristic subgroup of $G$ and hence $N=G$ because $N \ne (1)$ and $G$ is characteristically simple. The proof of Claim 2 is now complete.

We can now finish the proof of the Theorem. By Claim 2, $G=N_1 \times \cdots \times N_k$ for some normal subgroups $N_i$ such that $N_i \cong H$ for all $i.$ So the only thing left is to show that each $N_i$ is simple. Suppose that $K$ is a normal subgroup of some $N_i.$ Then $K$ is also a normal subgroup of $G$ because $N_i$ is a direct summand of $G.$ But $|K| \le |N_i|=|H|$ and so, by minimality of $|H|,$ we must have $|K|=1$ or $|K|=|H|=|N_i|.$ Hence either $K=(1)$ or $K=N_i,$ which proves that $N_i$ is simple. $\ \Box$

Corollary. Let $H$ be a minimal normal subgroup of a finite group $G.$ Then either $H$ is an elementary abelian $p$ -group or $H$ is the direct product of some isomorphic non-abelian simple groups.

Proof. By Example 2, $H$ is characteristically simple and hence, by the Theorem, $H$ is the direct product of some isomorphic simple groups $H_i.$ We are done if each $H_i$ is non-abelian. Suppose now that $H_j$ is abelian for some $j.$ Then $H_j \cong \mathbb{Z}_p$ for some prime $p,$ because $H_j$ is simple and abelian. Hence $H_i \cong H_j \cong \mathbb{Z}_p$ for all $i$ and thus $H$ is an elementary abelian $p$ -group. in this case. $\ \Box$

Note. The reference for the only if part of the Theorem and the Corollary is chapter 2, section 1, of the book Finite Groups by Daniel Gorenstein. Strangely, Gorenstein doesn’t even mention the if part of the Theorem in his book; the proof of that was added by me.

Exercise 1. Show that every characteristically simple group $G$ of order $n!$ is simple.
Hint. If $G$ is not simple, then, by the Theorem, $n!=|G|=|S|^k,$ for some simple group $S$ and some integer $k \ge 2.$ Show that the equation $n!=m^k$ has no solution for integers $m,n,k \ge 2,$ by, for example, proving that $n!$ has alway a prime divisor of exponent $1.$

Exercise 2. Show that a finite solvable group $G$ is characteristically simple if and only if it is an elementary abelian $p$ -group.
Hint. Since $G$ is solvable, $G' \ne G$ and so if $G$ is characteristically simple, then $G'=(1),$ i.e. $G$ is abelian. Now use the Theorem.

Exercise 3. Show that the symmetric group $S_n, \ n \ge 3,$ is not characteristically simple.
Hint. First Proof: The alternating group $A_n$ is a normal subgroup of $S_n$ and so $S_n$ is not simple hence not characteristically simple, by Exercise 1. Second Proof: $A_n$ is the unique subgroup of order $n!/2$ in $S_n$ and hence $f(A_n)=A_n$ for any automorphism $f$ of $S_n,$ i.e. $A_n$ is a characteristic subgroup of $S_n.$