Conjugates of a normal subgroup of a normal subgroup

Question

Let $A$ be a normal subgroup of $G$, and $B$ a normal subgroup of $A$.

I know that $B^g\le A^g=A$, namely the conjugates of $B$ in $G$ are still in $A$. But when I went further, I got stuck.

Are the conjugates of $B$ in $G$ still normal subgroups of $A$? In other words, my question is: if $B\trianglelefteq A\trianglelefteq G$, can we have $B^g\trianglelefteq A$ for all $g\in G$?

Any help is appreciated. Thank you!

Answer 1

The answer is Yes. In the following, fix a $g\in G$.

$B^g$ is normal in $A$ if and only if for every $b^g\in B^g$ and $a\in A$ holds $(b^g)^a\in B^g$. Since $A$ is normal in $G$, there is an $\bar a\in A$ so that $ag=g \bar a$ (basically, $\bar a=a^{g^{-1}}$, which I do not use for notational reasons). It follows

$$(b^g)^a=agbg^{-1}a^{-1}=(ag)b(ag)^{-1} = (g\bar a)b(g\bar a)^{-1}=g\bar a b \bar a^{-1}g^{-1} = (b^{\bar a})^g.$$

Since $B$ is normal in $A$, we have $b^{\bar a}\in B$ which shows that $(b^{\bar a})^g\in B^g$.