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 $G$.


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 $f(x)f(y)^{-1} = f(xy^{-1}) \in H$. $\blacksquare$

See also

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