I am new to characteristic subgroups, and have a theorem without a proof I don't understand yet. It says that if $A$ is characteristic in a group $G$ and $B$ is characteristic in $G$, then both $AB$ and $A\cap B$ are characteristic in $G$, i.e., characteristicity is closed under intersections and products. The book says the proof is trivial so it skips it, but I guess I'm not understanding the concept enough yet to see why this would be true.
Could anyone give me a quick intuitive reason why this is true and show me why the proof is "trivial"? I see online that people use this fact all the time, but can't seem to find any formal justification for it or proof of it. Thanks in advance for humoring me with a beginner's question...I'm sure it's not too hard once I see it.
$A$ si characteristic iff $\forall\psi\in\operatorname{Aut}(G),\,\psi(A)\subset A$.
For any map $\psi(A\cap B)\subset \psi(A)\cap\psi(B)$. In the case of characteristic subgroups we have for any automorphism
$$\psi(A\cap B)\subset \psi(A)\cap\psi(B)\subset A\cap B$$
And this proves the closedness under intersection
Do you feel like proving it for the product ?