Recall that an automorphism $\sigma$ of a group $G$ is a group isomorphism from $G$ to itself.
The set of all automorphism of $G$ is denoted by $\Aut(G)$.

Proof.

(a) If $\sigma(H) \subset H$ for all $\sigma \in \Aut(G)$, then $H$ is characteristic in $G$

Since $\sigma$ is an automorphism, the inverse $\sigma^{-1}$ is also an automorphism of $G$.
Hence, we have
\[\sigma^{-1}(H)\subset H\]
by the assumption.

Applying $\sigma$, we have
\[\sigma\sigma^{-1}(H) \subset \sigma(H).\]
Then we obtain
\begin{align*}
H&=\sigma \sigma^{-1}(H)\subset \sigma(H)\subset H.
\end{align*}

Since the both ends are $H$, the inclusion is in fact the equality.
Thus, we obtain
\[\sigma(H)=H,\]
and the subgroup $H$ is characteristic in the group $G$.

(b) The center $Z(G)$ of $G$ is characteristic in $G$

By part (a), it suffices to prove that $\sigma(Z(G)) \subset Z(G)$ for every automorphism $\sigma \in \Aut(G)$ of $G$.

Let $x\in \sigma(Z(G))$. Then there exists $y \in Z(G)$ such that $x=\sigma(y)$.
To show that $x \in Z(G)$, consider an arbitrary $g \in G$.
Then since $\sigma$ is an automorphism, we have $G=\sigma(G)$.
Thus there exists $g’$ such that $g=\sigma(g’)$.

We have
\begin{align*}
xg &=\sigma(y)\sigma(g’)\\
&=\sigma(yg’) && \text{ (since $\sigma$ is a homomorphism)}\\
&=\sigma(g’y) && \text{ (since $y \in Z(G)$)}\\
&=\sigma(g’)\sigma(y) && \text{ (since $\sigma$ is a homomorphism)}\\
&=gx.
\end{align*}
Since this is true for all $g \in G$, it follows that $x \in Z(G)$, and thus
\[\sigma(Z(G)) \subset Z(G).\]
This completes the proof.

Comment.

In some textbook, a subgroup $H$ of $G$ is said to be characteristic in $G$ if $\sigma(H) \subset H$ for all $\sigma \in \Aut(G)$.
Problem (a) implies that our definition of characteristic and this alternative definition are in fact equivalent.

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