Characteristicity is quotient-transitive

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 $$G$$.

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 $$\sigma'$$.

Then, $$\sigma'$$ is an automorphism of $$G/H$$. Since $$K/H$$ is characteristic in $$G/H$$, $$\sigma'(K/H) = K/H$$. Thus, for any $$g \in K$$, $$\sigma'(gH) \in K/H$$, 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.