Suppose that $G_{1}$ and $G_{2}$ are groups, and there is a short exact sequence $$1 \rightarrow N \rightarrow G_{1} \xrightarrow{\alpha} G_{2} \rightarrow 1.$$ Let $H$ be a subgroup in $G_{2}$ (not normal) and suppose that $$G_{2}= Hx_{1}\dot{\cup}\cdots \dot{\cup} Hx_{n}.$$
Assume that $y_{1},\cdots,y_{n}$ are elements in $G_{1}$ such that $\alpha(y_{j})=x_{j}$ for all $j\in \{1,\cdots,n\}$.
Is it true that $G_{1}=\langle N, H \rangle y_{1}\dot{\cup}\cdots \dot{\cup} \langle N,H \rangle y_{n}$?
My attempt: Let $x$ be an element in $G_{1}$. Then, since $G_{1}/N$ is isomorphic to $G_{2}$, $Nx= N z y_{j}$ for some $j$ and $z$ is a lift of an element of $H$. Then, $x=n z y_{j}$ where $n$ is an element in $N$.
Am I right? Thanks in advance!
Yes, this is true. By the correspondence theorem $H$ corresponds to a subgroup of $G_1$ that contains $N$, namely, $\mathcal{H}=a^{-1}(H)$ (which is not exactly $\langle N,H\rangle$ as that does not really make sense: $H$ is not a subgroup of $G_2$, even if you want to view $N$ as a subgroup of $G_2$).
Now, note that $y_j^{-1}y_i\in \mathcal{H}$ if and only if $a(y_j^{-1}y_i)\in H$ if and only if $a(y_j)^{-1}a(y_i)\in H$, if and only if $x_j^{-1}x_i\in H$, if and only if $j=i$ (by the choice of the $x_k$). Thus, the $y_k$ represent distinct cosets of $\mathcal{H}$ in $G_2$. And given any $g\in G_2$, there exists $x_i$ such that $a(g)\in x_iH$, hence $g\in a^{-1}(x_iH) = y_i\mathcal{H}$, so this is a complete set of representatives.