I need an example to show that the following statement may be false:
For $i = 1, 2$, let $H_i$ be a normal subgroup of $G_i$
If $G_1\cong G_2$ and $G_1/H_1 \cong G_2/H_2$ then $H_1 \cong H_2$.
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A quotient group is denoted $G/H$, not $G\backslash {H}$. – Bernard Oct 09 '19 at 17:56
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Knowing that any subgroup of index $2$ is normal (Subgroup of index 2 is Normal), you can just pick a group with two non-isomorphic subgroups of index $2$.
For example, take the dihedral group $D_4=\{e,\rho, \rho^2, \rho^3, \sigma, \rho\sigma, \rho^2\sigma, \rho^3\sigma\}$ and its subgroups $<\rho>=\{e, \rho, \rho^2, \rho^3\}$ and $<\rho^2, \sigma>=\{e, \rho^2, \sigma, \rho^2\sigma\}$: the former subgroup is cyclic, the latter is not.