Let $G$ is a group and $N$ is normal in $G$ and $M_1$ and $M_2$ are subgroups which contain $N$ such that $M_1/N= M_2/N$. Can we deduce that $M_1$ and $M_2$ are isomorphic?

Thank you for hints.

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    $\begingroup$ Why did you use $=$ instead of $\cong$? What do you think about the proof? $\endgroup$ – mrs Jan 13 '13 at 15:28
  • 1
    $\begingroup$ Are you sure that you want $M_1/N=M_2/N$ and not $M_1/N\cong M_2/N$ $\endgroup$ – Amr Jan 13 '13 at 15:31
  • $\begingroup$ If $M_1/N=M_2/N$ then $M_1=M_2$. $\endgroup$ – user26857 Jan 13 '13 at 15:35

A very short way to see this: Any subgroup has a unique coset which is a subgroup (the subgroup itself) so if two subgroups have the same cosets they must be equal.


(assuming = means "are isomorphic") I don't think so, let's just work with abelian groups. Consider $\mathbb{Z}_2 \oplus \mathbb{Z}_4$ with subgroups $\mathbb{Z}_4$ generated by $(0,1)$ and $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ with generators $(1,0)$ and $(0,2)$, and consider the span of $(0,2)$.


If we consider $*=*$ as you noted somewhre above is as $*\cong*$, then the claim looks fine. In fact @YACP commented right. Because if $m_1\in M_1$ so $$m_1N\in\frac{M1}{N}\cong\frac{M_2}{N}$$ so $m_1N\in\frac{M_2}{N}$ and this means that for $m_2\in M_2$. we have $m_1N=m_2N$ or $m_2^{-1}m_1\in N$. But $N\subset M_2$ so $m_2^{-1}m_1\in M_2$. What does this mean?

  • $\begingroup$ Or a simpler argument, the underlying set of $M_i/N$ is a partition of $M_i$. The union of the cosets is equal to $M_i$, but the collections are the same so the union is the same. $\endgroup$ – Asaf Karagila Jan 13 '13 at 16:11
  • $\begingroup$ @AsafKaragila: Yes. It seems I walked a longer zig-zag way. Thanks Asaf. $\endgroup$ – mrs Jan 13 '13 at 16:26
  • $\begingroup$ Helpful!! +1 ${}{}$ $\endgroup$ – Namaste Feb 18 '13 at 0:13

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