Does $ \frac{G}{H}\simeq \frac{G}{K}$ $\Rightarrow$ $H\simeq K$?

Question

Does $\displaystyle \frac{G}{H}$ $\simeq$ $\displaystyle \frac{G}{K}$ $\Rightarrow$ $H$ $\simeq$ $K$?

I think it's true but I am having trouble demonstrating it.

If $H$ and $K$ are subgroup of a group $G$ such that $\displaystyle \frac{G}{H}$ is isomorphic a $\displaystyle \frac{G}{K}$, then $H$ is isomorphic a $K$.

Consider the dihedral group of order $8$ for $G$ and think about its normal subgroups of order $4.$ — Geoff Robinson, Jun 25 '14 at 12:46

score 6 · Accepted Answer · edited Apr 13 '17 at 12:20

6

No. Consider, for example, $\mathbb{Z}_2\times\mathbb{Z}_4$.

For more extreme examples, see here. These examples are of infinite groups where $H$ is trivial but $K$ is non-trivial. Such groups are called non-Hopfian.

edited Apr 13 '17 at 12:20

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answered Jun 25 '14 at 12:50

user1729

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Does $ \frac{G}{H}\simeq \frac{G}{K}$ $\Rightarrow$ $H\simeq K$?

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