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Suppose $G$ is a finite group and $N$ is a normal subgroup of $G$. Then subgroups of $G/N$ are of the form $A/G$ for $A\le G$. But how does the normal subgroups of $G/N$ look like? Is it true that $A \unlhd G \iff A/N \unlhd G/N$ provided $N$ is a subgroup of $A$?

I can see that if $A\unlhd G$ and $N\unlhd A$ then $A/N \unlhd G/N$.

If I then assume that $A/N \unlhd G/N$ and take $xN \in A/N$ and $yN \in G/N$ then $(xN)^{(yN)} = (x^y)N \in A/N$. Does this imply that $x^y \in A$?

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You're definitely on the right track. You need to show that $xN \in A/N \implies x \in A$.

Some hints to think about:

$1)$ What does it mean for $xN$ to be in $A/N$?

$2)$What does it mean for two cosets to be the same?

Hopefully this should help you out.

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  • $\begingroup$ $xN \in A/N \iff (xN)A/N = A/N$, right? But I am not sure how this helps me... $\endgroup$ – user129954 Jul 10 '14 at 10:32
  • $\begingroup$ @user129954 That's certainly true, but I was thinking more along the lines of $\exists a \in A$ such that $xN=aN$, aince $A/N = \{aN:a\in A\}$ $\endgroup$ – Tom Oldfield Jul 10 '14 at 11:26
  • $\begingroup$ I am sorry, I still don't see it. If $xN = aN$ then $a^{-1}xN = N$ so $a^{-1}x\in N$. And I know that $a^{-1}\in A$. But how does it help me? $\endgroup$ – user129954 Jul 24 '14 at 11:21
  • $\begingroup$ No I am sorry, it is obvious then $a^{-1}x\in A$ so $x\in A$ as $a^{-1}\in A$. $\endgroup$ – user129954 Jul 24 '14 at 11:24
  • $\begingroup$ @user129954 Yes, that's right. $\endgroup$ – Tom Oldfield Jul 24 '14 at 18:47

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