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Suppose $\sigma:H\to G$ is a group homomorphism with $G,H$ finite and $\mathrm{ker}(\sigma)=Z(H)$.

If $\sigma$ restricts to isomorphisms $$\sigma|_{H_1}:H_1\to G_1$$ $$\sigma|_{H_2}:H_2\to G_2$$ where each $H_i$ is cyclic of coprime order then is $\sigma|_{\langle H_1,H_2\rangle}:\langle H_1,H_2\rangle\to\langle G_1,G_2\rangle$ necessarily an isomorphism?

Some intuition:

I feel like this should have either an easy proof or an obvious counterexample and perhaps doesn't require all the hypotheses, perhaps for example we may only require that $G_1\cap G_2$ is trivial or that $H_1,H_2$ are simple.

If $H_1$ or $H_2$ is normal in $G$ then $G_1\cap G_2$ trivial suffices as $\langle H_1,H_2\rangle=H_1H_2$ and $|H_1H_2|=|H_1||H_2|=|G_1||G_2|=|G_1G_2|$ so surjectivity of $\sigma|_{\langle H_1,H_2\rangle}$ implies bijectivity so $\sigma|_{\langle H_1,H_2\rangle}$ is an isomorphism. It is not clear that this should generalise to the case $H_1,H_2$ are not normal in $G$.

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Here's a counterexample. Let $K$ be a finite field and let $a\in K^\times$ be nontrivial and have odd order (this is possible as long as $|K|-1$ is not a power of $2$). Let $H=GL_2(K)$, $h_1=\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}$, and $h_2=\begin{pmatrix}a & 0 \\ 0 & 1\end{pmatrix}$. Then the order of $h_1$ is $2$ and the order of $h_2$ is the order of $a$, and the cyclic subgroups generated by both have trivial intersection with the center. So setting $G=H/Z(H)$ with $\sigma$ the quotient map, the subgroups $H_i$ generated by $h_i$ satisfy your hypotheses. But notice that $h_2h_1h_2h_1=\begin{pmatrix} a & 0 \\ 0 & a\end{pmatrix}$ which is a nontrivial element of the center. So $\sigma$ is not injective on $\langle H_1,H_2\rangle$.

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