Characteristicity is quotient-transitive

This article gives the statement, and possibly proof, of a subgroup property (i.e., characteristic subgroup) satisfying a subgroup metaproperty (i.e., quotient-transitive subgroup)
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Statement

Property-theoretic statement

The subgroup property of being a characteristic subgroup satisfies the subgroup metaproperty of being quotient-transitive.

Statement with symbols

Suppose $H\le K\le G$ are subgroups such that $H$ is a characteristic subgroup of $G$, and $K/H$ is a characteristic subgroup of $G/H$. Then, $K$ is a characteristic subgroup of $G$.

Related facts

Similar facts

• Normality is quotient-transitive
• Full invariance is quotient-transitive
• Strict characteristicity is quotient-transitive

Proof

Given: A group $G$, subgroups $H\le K\le G$ such that $H$ is characteristic in $G$, and $K/H$ is characteristic in $G/H$

To prove: $K$ is characteristic in $G$

Proof: We pick any automorphism $\sigma$ of $G$, and want to show that $\sigma (K)=K$. For this, first observe that $\sigma (H)=H$, so $\sigma$ induces an automorphism on the quotient $G/H$, by the rule $gH\mapsto \sigma (g)H$. Call this automorphism ${\sigma }^{\prime }$.

Then, ${\sigma }^{\prime }$ is an automorphism of $G/H$. Since $K/H$ is characteristic in $G/H$, ${\sigma }^{\prime }(K/H)=K/H$. Thus, for any $g\in K$, ${\sigma }^{\prime }(gH)\in K/H$, and hence, unwrapping the definition, $\sigma (g)\in K$. Thus, $\sigma (K)\subset K$. Since the same holds for ${\sigma }^{-1}$, we conclude that $\sigma (K)=K$, completing the proof.