1
$\begingroup$

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?

Improve this question
$\endgroup$
  • $\begingroup$ @Watson the dupe is slightly problematic as the accepted answer actually seems to use equality not isomorphy. $\endgroup$ – quid♦ Aug 22 '16 at 21:40
  • $\begingroup$ @Brian please see my comment above. $\endgroup$ – quid♦ Aug 22 '16 at 21:40
  • 2
    $\begingroup$ @quid: Ok, I agree. But isn't this answer fine: math.stackexchange.com/q/276976? What do you think? $\endgroup$ – Watson Aug 22 '16 at 21:41
  • 1
    $\begingroup$ @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. $\endgroup$ – quid♦ Aug 22 '16 at 21:45
  • 2
    $\begingroup$ 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. $\endgroup$ – arctic tern Aug 22 '16 at 23:57
2
$\begingroup$

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 $p=2$.)

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.