For a subgroups $H_1$ and $H_2$ of $G$ and a normal subgroup $N$ of $G$: $H_1/N \cong H_2/N \Rightarrow H_1 \cong H_2$. Is it true?

I need to understand what can we deduce from knowing that:

Let $G$ be a group, $H_1, H_2$ its subgroups and $N$ it's normal subgroup contained in both $H_1$ and $H_2$. Assume

$H_1/N \cong H_2/N$

Can we really say that then $H_1 \cong H_2$? Or anything else related?

• @Watson the dupe is slightly problematic as the accepted answer actually seems to use equality not isomorphy. – quid Aug 22 '16 at 21:40
• @Brian please see my comment above. – quid Aug 22 '16 at 21:40
• @quid: Ok, I agree. But isn't this answer fine: math.stackexchange.com/q/276976? What do you think? – Watson Aug 22 '16 at 21:41
• @Watson yes, that answer would be a correct answer to this question. But the question there asks for something different as written and also according to its accepted answer. I find this way too confusing to dupe-close. – quid Aug 22 '16 at 21:45
• Watson is right, the answer over there by uncookedfalcon (which is the one Watson specifically linked to) also answers your question. I can reproduce it here in the comments for convenience: Let $G=\Bbb Z_2\times \Bbb Z_4$ and $H_1=\langle (0,1)\rangle\cong\Bbb Z_4$ and $H_2=\langle (1,0),(0,2)\rangle\cong\Bbb Z_2\times\Bbb Z_2$. Then $H_1/N\cong\Bbb Z_2\cong H_2/N$ but $H_1\not\cong H_2$ are not isomorphic. – arctic tern Aug 22 '16 at 23:57

The groups are not necessarily isomorphic.

A good strategy to approach such a question is to consider the simplest case first.

Thus, to make things simple let us consider $N$ is a cyclic group of prime order $p$ sitting inside a to be specified group.

For any subgroup $H$ of order $p^2$ that contains $N$, we'll have $H/N$ is of order $p$ and thus also cyclic.

Thus if what you ask about were true then all subgroups of order $p^2$ that contain $N$ would need to be isomporphic.

Is there a good reason for this? Not really. There are two types of groups of order $p^2$, the cyclic one of order $p^2$ and the direct product of two cyclic one of order $p$.

Can we find a group that contains both these types of subgroups and have non-empty intersection?

Let us take the direct product of a cyclic group of order $p^2$ with a cyclic group of order $p$. Then we certainly have cyclic subgroups of order $p^2$ and also a direct product of cyclic groups of order $p$.

For example, if we write the group elements as pairs and use additive notation, first coordinate order $p^2$ second $p$, then the group generated by $(1,0)$ that is $H_1=\{(0,0),(1,0),(2,0), \dots, (p^2 -1,0)\}$ is cyclic of order $p^2$, the one generated by $(p,0)$ and $(0,1)$ is a product of two cyclic groups of order $p$, call it $H_2$; and the intersection is $N=\{(0,0), (p,0), (2p,0), (3p,0), \dots, (p(p-1), 0)\}$ cyclic of order $p$.

Then $H_1/N$ and $H_2/N$ are isomorphic (both cyclic of order $p$) while $H_1$ and $H_2$ are not.

Various other examples of this form could be given. (The example discussed in comments is the special case $p=2$.)