Conjugacy of direct product of groups

Question

All groups are finite here. Let $G=\langle a,b \rangle \times H$, $a,b$ commute, $G_{1}=\langle a \rangle \times H_{1}$ and $G_{2}=\langle a \rangle \times H_{2}$ where $H_{1}$ and $H_{2}$ are subgroups of $H$. If $G_{1}$ and $G_{2}$ are conjugate, are $H_{1}$ and $H_{2}$ conjugate?

I'm aware that $s(M\times N)s^{-1}=sMs^{-1} \times sNs^{-1}$. So there is $g\in G$ s.t. $g(\langle a \rangle \times H_{1})g^{-1}=g \langle a \rangle g^{-1} \times gH_{1}g^{-1} = \langle a \rangle \times H_{2}$. Can we deduce from this that $g \langle a \rangle g^{-1} = \langle a \rangle$ and $gH_{1}g^{-1} = H_{2}$? It feels a bit off to me. Thx

Answer 1

Sorry, I misread the question initially, partly because the element $b$ seems to be irrelevant to the question.

We have $a,b \in Z(G)$, so if the subgroups $\langle a,H_1\rangle$ and $\langle a,H_2\rangle$ are conjugate in $G$, then they are conjugate in $H$, and so the answer is yes.

There is no need to assume that all groups are finite.

Answer 2

It can also be done more explicitly (even though it's not elegant at all).

We want to see there is an element $g \in G$ such that $H_1 = g H_2 g^{-1}$. By the symmetry of the problem, it suffices to show that $g H_2 g^{-1} \subseteq H_1$ for some element $g \in G$.

By hypothesis, since $G_1$ and $G_2$ are conjugate there is an element $(\tau,g) \in \langle a, b \rangle \times H = G$ such that $G_1 = (\tau, g) G_2 (\tau, g)^{-1}$.

Pick an element $h \in H_2$. We show that $ghg^{-1} \in H_1$; this will complete the proof. By the definition of $(\tau, g)$, we know that $(\tau, g) (1, h) (\tau, g)^{-1} \in G_1$, i.e. $(\tau\tau^{-1}, ghg^{-1}) \in G_1$. By the definition of $G_1$, this means that $ghg^{-1} \in H_1$. This completes the proof.