# $M_1$ and $M_2$ are subgroups and $M_1/N=M_2/N$. Is $M_1\cong M_2$?

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.

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

(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?
• 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. – Asaf Karagila Jan 13 '13 at 16:11
• Helpful!! +1 ${}{}$ – Namaste Feb 18 '13 at 0:13