Decomposition of a quotient group into products Let $G$ and $H$ be groups and $L$ a normal subgroup of $G\times H$. Suppose $L\cong G'$, where $G'$ is a normal subgroup of $G$. Under what conditions the following isomorphism hold?
$(G\times H)/L\cong (G\times H)/(G'\times \{e\})\cong (G/G')\times H$
Satisfying the required conditions, what is the explicit isomorphism between $(G\times H)/L$ and $(G/G')\times H$?
 A: A you probably guessed, this is not true in general.
Let $p$ be an odd prime, and let $G = G_1$, $H = G_2$ be two copies of the nonabelian group of order $p^{3}$ and exponent $p^{2}$,
$$
G_{i} = \langle a_{i}^{p^{2}} = b_{i}^{p} = 1, [a_{i}, {b_{i}}] = a_{i}^{p}\rangle.
$$
Consider the two normal subgroups of order $p$ of $G_1 \times G_2$ given by $L =\langle a_1^{p} a_2^{-p} \rangle$ and $G' = \langle a_{1}^{p} \rangle \le G_{1}$.
Compare $\Omega_{1}((G_1 \times G_2) / G')$ and $\Omega_{1}((G_1 \times G_2) / L)$, the set of elements of order $p$, which is a subgroup in this case. In both cases it has order $p^{4}$, but in the first case it is abelian, while in the second case the cosets of $a_{1} a_{2}^{-1}$ and $b_{1}$ do not commute, as the commutator
$$
[a_{1} a_{2}^{-1}, b_{1}] = a_{1}^{p}
$$ 
does not belong to $L$.
PS As an additional condition under which you have isomorphism, I can only think at the moment of the existence of an automorphism of the direct product which takes $G'$ to $L$.
