# $H_1\triangleleft G_1$, $H_2\triangleleft G_2$, $H_1\cong H_2$ and $G_1/H_1\cong G_2/H_2 \nRightarrow G_1\cong G_2$

Find a counterexample to show that if $G_1$ and $G_2$ groups,

$H_1\triangleleft G_1$, $H_2\triangleleft G_2$, $H_1\cong H_2$ and $G_1/H_1\cong G_2/H_2 \nRightarrow G_1\cong G_2$

I tried but I did not have success, I believe that these groups are infinite.

Let $G_1$ and $G_2$ be two nonisomorphic groups of order $4$. (What are the only two possibilities?)

$G_1$ and $G_2$ each have a subgroup of order $2$, which is isomorphic to $\mathbb{Z}/2\mathbb{Z}$ since that is the only group of order $2$. Since $G_1$ and $G_2$ are both abelian, all subgroups are normal, and the quotient groups are also of order $2$ (why?), so the quotient groups are also nonisomorphic.

The Group Extension Problem asks: What are all the possibilities for $G$ if we know a normal subgroup $H$ and $G/H$? For abelian groups, the problem can be solved using homological algebra, but it is wide open in general.

• You are right, your counterexample is even more simple.
– Tile
Commented Dec 17, 2012 at 15:48

There have already been nice answers but I will give mine example.

Take $H_1=H_2=C_3$, the cyclic group of order $3$. Take $G_1=D_6$ and $G_2=C_6$, then one sees both $G_1/H_1$ and $G_2/H_2$ are $C_2$, but obviously $G_1$ and $G_2$ are not isomorphic, one being abelian while the other is not.

The standard counter example to that implication is the quaternion group $Q_8$ and dihedral group $D_4$.

Both groups have order $2^3=8$, so that every maximal group (i.e. one of order 4) is normal. The cyclic group of order 4 is contained in both groups and the quotient has order 2 in both cases. So all assertions are satisfied but $D_4\ncong Q_8$, of course.