TITLE: If $N$ is a normal subgroup of a group $G$, and $M$ is a characteristic subgroup of $N$, then $M$ is a normal subgroup of G.
QUESTION [5 upvotes]: Prove that:
If $N$ is a normal subgroup of a group $G$, and $M$ is a characteristic subgroup
of $N$, then $M$ is a normal subgroup of G.
Here what I am seeing is that $M$ is normal in $N$ and $N$ normal in $G$. But normality is not transitive property. So how to go?

REPLY [2 votes]: Let f be an inner automorphism of G such that f : G $\to$ G . Then, the restriction of f to N is an automorphism of N since N is a normal subgroup of G, as previously stated. Here is where things get cleared up: Since M is a characteristic subgroup of N, it is invariant under f since all characteristic subgroups are invariant under all automorphisms. We can say this for our inner automorphism f since Inn(G) $\subset$ Aut(G). Since a subgroup that is invariant under all inner automorphisms is normal, M must be normal.
Short hand: M is characteristic $\Rightarrow$ M is invariant under all automorphisms $\Rightarrow$ M is normal. In our case, the automorphism we are talking about is f that holds from N to N since N is normal in G.