The commutator of the product of 2 groups

Question

Let $G=H\times A$ be the internal product of $A$ and $H$,

I proved that if "$A$ is abelian then $G′=H′$ "

But if we were not given that $A$ is abelian is $G′=H′\times A′$ ?

Answer 1

Let's consider a typical generator of $G'$, by picking any two elements of $G$, namely $g_1 = (h_1,a_1)$ and $g_2 = (h_2,a_2)$, and forming their commutator: \begin{align*} [g_1,g_2] &= g_1 \cdot g_2 \cdot g_1^{-1} \cdot g_2^{-1} \\ &= (h_1,a_1) \cdot (h_2, a_2) \cdot (h_1^{-1},a_1^{-1}) \cdot (h_2^{-1},a_2^{-1}) \\ &= (h_1 \cdot h_2 \cdot h_1^{-1} \cdot h_2^{-1}, a_1 \cdot a_2 \cdot a_1^{-1} \cdot a_2^{-1} )\\ &= ([h_1,h_2],[a_1,a_2]) \\ &\in H' \times A' \end{align*} Since elements of the form $[g_1,g_2]$ are generators of $G'$, this proves that $G' < H' \times A'$. The opposite inclusion should be obvious. So yes, $G' = H' \times A'$.