Characteristic subgroup

A characteristic subgroup of a group $G$ is a subgroup of $G$ that is stable under every automorphism on $G$ . Since the map $y \mapsto xyx^{-1}$ is an automorphism (specifically, an inner automorphism) on $G$ , for every $x$ , it follows that every characteristic subgroup of $G$ is also a normal subgroup of $G$ .

Examples

Every group is a characteristic subgroup of itself; a group's trivial subgroup is characteristic.

Let $n$ be a natural number that divides the order of $G$ . Then the set of elements $a$ of $G$ for which $\text{ord}(a)$ divides $n$ is a characteristic subgroup of $G$ . Since every subgroup of a cyclic group is cyclic, it follows that every subgroup of a cyclic group is a characteristic group.

In general, though, not every cyclic group of an Abelian group is characteristic. For instance, the Klein 4-group has no non-trivial characteristic subgroups, since any permutation of its non-identity elements is an automorphism. For an odd prime $p$ , the group $G=(\mathbb{Z}/p\mathbb{Z})^2$ has no nontrivial characteristic subgroups either. Indeed, for any $\alpha, \beta$ relatively prime to $p$ , the mapping $(a,b) \mapsto (\alpha a + \beta b, \alpha a - \beta b)$ is an automorphism, as is the mapping $(a,b) \mapsto (b,a)$ . Thus if $(a,0)$ ( $a \neq 0$ ) is a member of a characteristic subgroup, then so is $(0,a)$ , and these two evidently generate $G$ ; and if $(a,b)$ ( $a,b \neq 0$ ) is an element of a characteristic subroup, then setting $\alpha = b, \beta = a$ , we see that $(2ab,0)$ is an element of this characteristic subgroup; therefore so is all of $G$ .

This idea also shows that $(\mathbb{Z}/p\mathbb{Z})^n$ has no non-trivial characteristic subgroups, for any natural number $n$ . In fact, the characteristic subgroups of $(\mathbb{Z}/p^k\mathbb{Z})^n$ are of the form $(p^a \mathbb{Z}/p^k\mathbb{Z})^n$ , for some integer $a<k$ .

Characteristic Subgroups of Normal Subgroups

Theorem 1. Let $G$ be a group, and let $H$ be a normal subgroup of $G$ . Let $K$ be a characteristic subgroup of $H$ . Then $K$ is a normal subgroup of $G$ ; furthermore, if $H$ is a characteristic subgroup of $G$ , then so is $K$ .

Proof. Let $f$ be an (inner) automorphism on $G$ . Then its restriction to $H$ is an automorphism on $H$ , so $K$ is characteristic (normal) in $G$ if and only if $H$ is. $\blacksquare$

Theorem 2. An equivalence relation $\mathcal{R}(x,y)$ is compatible with the group law on $G$ and every automorphism on $G$ if and only if $\mathcal{R}(x,y)$ is equivalent to $xy^{-1} \in H$ , for some characteristic subgroup $H$ of $G$ .

Proof. Evidently, $\mathcal{R}(x,y)$ must be of the form $xy^{-1} \in H$ , where $H$ is the set of elements equivalent to $e$ under $\mathcal{R}$ . For any $a \in H$ and any automorphism $f$ on $G$ , $f(a) \equiv f(e) \equiv e \pmod{H}$ , so $H$ is a characteristic subgroup.

On the other hand, if $H$ is a characteristic subgroup, the relation $xy^{-1} \in H$ is compatible with the group law; it also implies $f(x)f(y)^{-1} = f(xy^{-1}) \in H$ . $\blacksquare$

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Characteristic subgroup

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Characteristic Subgroups of Normal Subgroups

See also