# Are these subgroups isomorphics? [closed]

Let $H_1\leq G_1$ and $H_2\leq G_2$. If $G_1\simeq G_2$ and $G_1/H_1\simeq G_2/H_2$ is $H_1\simeq H_2$

thanks

• Probably related.
– Vim
Commented May 23, 2016 at 0:04
• The title reminds me of Gollum talking from lord of the rings. Commented May 23, 2016 at 3:43

## 4 Answers

This is not even true for finite abelian groups.

Let $C_n$ denote the cyclic group of order $n$. If $G=C_4 \times C_2$, then $G/(C_2\times C_2) \cong C_2 \cong G/(C_4 \times 1)$, but $C_2\times C_2\not\cong C_4$.

The groups $S_3$ and the cyclic group of order 6 are non-isomorphic. But both have cyclic subgroups of order 3 as normal subgroups (in the former it is $A_3$).

The quotients in both the cases is the cyclic group of order 2.

The moral from all these examples is: a group is obtained with many subgroups assembled in many different ways. So it is possible to assemble same pieces in different manners to obtain different things. The concept of group extensions tries to formalize this idea.

No, take $G=\mathbb{Z}^{\mathbb{N}}$ the product of $\mathbb{N}$-copies of $\mathbb{Z}$, the quotient of $G$ by $H_1=\{(x,y,0...); x,y\in \mathbb{Z}\}$ is isomorphic to the quotient of $G$ by $H_2=\{(x,0,...); x\in \mathbb{Z}\}$. The group $H_1$ is isomorphic to $\mathbb{Z}^2$ and $H_1$ is isomorphic to $\mathbb{Z}$.

• But, why $G/H_1$ is isomorphic to $G/H_2$ Commented May 23, 2016 at 0:01
• Because they are isomorphic to $Z^N$ Commented May 23, 2016 at 0:02

No. Exemple $G_1=G_2= S_3 \times \mathbb{Z}/6\mathbb{Z}$, the group quotient respectively with the no isomorphic subgroup $S_3 \times \mathbb{Z}/3\mathbb{Z}$ and $A_3 \times \mathbb{Z}/6\mathbb{Z}$ is the same $\mathbb{Z}/2\mathbb{Z}$. There is others similar question in this website good luck

• S3×Z/3Z and A3×Z/6Z are not isomorphic subgroup of S3×Z/6Z, and are normal, so by passing to quotient we have the same group qoutient Z/2Z Commented May 23, 2016 at 10:21