Do a subgroup and quotient determine the original group? I am wondering whether there is a counterexample which shows that subgroups and quotients don't determine the group.
More precisely, suppose there are two groups $G_1, G_2$ such that all of their proper non-trivial normal subgroups are 1-1 corresponding and if $1<H_1 < G_1, 1<H_2 < G_2$ are those proper normal subgroups that they correspond, then $H_1 \simeq H_2$, and $G_1 / H_1 \simeq G_2/H_2$.  (Here $\simeq$ means isomorphic.)
Then $G_1 \simeq G_2$?
I guess it might be not true in general, but I don't know any nontrivial counterexample except the pair $(\mathbb{Z}_p, \mathbb{Z}_q)$. 
Any comments on this will be highly appreciated!
 A: Smallest counterexample: $G_1=C_4$ generated by $x$ and $G_2=C_2 \times C_2$. Take $H_1=\langle x^2 \rangle$ and $H_2=C_2 \times \{1\}$.
A: There is a  counterexample of order $605 = 11^2 \times 5$ with the structure $11^2:5$. Let
$$G = \langle x,y,z \mid x^{11}=y^{11}=z^5=1, xy=yx, x^z=x^4, y^z=y^5 \rangle,$$
$$H = \langle x,y,z \mid x^{11}=y^{11}=z^5=1, xy=yx, x^z=x^4, y^z=y^3 \rangle.$$
These are $\mathtt{SmallGroup}(605,5)$ and $\mathtt{SmallGroup}(606,6)$ in the small groups database.
They both have $121$ conjugate subgroups of order $5$, two normal subgroups of order $11$ with nonabelian quotients of order $55$, $10$ non-normal subgroups of order $11$, $22$ non-normal subgroups of order $55$, and one normal subgroup of order $121$.
A: There are two (1) groups of order $4$.  Each has a normal subgroup of order $2$ so the quotient groups are also order $2$.  There is only one (1) group of order $2$ so the subgroups are isomorphic and the quotient groups are isomorphic.
(1) Up to isomorphism.
See The extension problem (Wikipedia)
