Subgroup of semidirect product

Question

Let $G$ be a semidirect product of a normal subgroup $A$ with a subgroup $B$ and Let $H$ be a subgroup of $G$ such that $H\cap A$ is trivial. Is it true that $H$ is contained in a conjugate of $B$ ? How can I prove that?

Answer 1

Alas, it not true.

Possibly the simplest example is the dihedral group of order $8$, $$ G = \langle a, b : a^4 = 1, b^2 = 1, b^{-1} a b = a^{-1} \rangle, $$ with $A = \langle a \rangle$ and $B = \langle b \rangle$.

You can see that $H = \langle b a \rangle$, a subgroup of order $2$ which intersects $A$ trivially, is not conjugate to $B$.

This is because the centralizer of $b$ is $\langle b, a^{2} \rangle$, of order $4$. Thus $b$ has two conjugates, which are $b$ itself and $b^{a} = a^{-1} b a = b (b^{-1} a^{-1} b) a = b a^{2}$.