Is the quotient of a normal subgroup necessarily normal? Let $N,M$ be normal subgroups of $G$.
Is it necessarily true that $M/N$ is normal in $G/N$?
The third isomorphism says the converse to be true, namely that if a group in $G/N$ is normal, then it is of the form $M/N$ for some normal subgroup $M$ of $G$.

My attempt at proving this as true:


*

*$G/N$ consists of cosets of $N$. i.e. a group element looks like $gN$
for $g \in G$.

*$M/N$ consists of cosets of $N$. i.e. a group element looks like $mN$
for $m \in M$.

*$M/N$ as a set may be expressed as $\{mN : m \in M\}$

*If we conjugate everything in $M/N$ by something in $G/N$ we get $\{gN \cdot mN : m \in M\} = \{g^{-1}mgN : m \in M\} = \{mN : m \in M\}$ 


Is this correct? I feel it is, but I'm suspicious due to the fact that this isn't listed in the isomorphism theorems on Wikipedia...
 A: Suppose $N$ and $M$ are normal subgroups of $G$.  If $N \subseteq M$, it's easy to see that $M/N$ is a normal subgroup of $G/N$.  But you are not assuming $N \subseteq M$, so this is just a special case.
Do not assume that $N \subseteq M$.  Let $H = NM = \{ nm : n \in N, m \in M \}$.  Note that $H = MN$.  Since $N$ and $M$ are both normal subgroups of $G$, so is $H$.
The set $M/N$ of cosets $mN : m \in M$ is a subgroup of $G/N$.  It is equal to $H/N$, which is normal in $G/N$ by the special case mentioned above.  So $M/N$ is normal in $G/N$ even if $N \not\subseteq M$.
Another way: This is basically what you did  in your post, I'm just going to state it in a different way.  If $\pi: G \rightarrow G'$ is a surjective group homomorphism, and $M$ is normal in $G$, then $\pi(M)$ is a normal subgroup of $G'$.  Again pretty much the same as your case: if $N = \textrm{Ker}(\pi)$, then you can identify $G'$ with $G/N$ via the first isomorphism theorem, under which $\pi(M)$ gets identified with $M/N$.
A: Let $gN \in G/N$ and $hN \in M/N$.
In particular, $g \in G$ and $h \in M$.
So, $ghg^{-1} \in M$.
Then, $(gN)(hN)(gN)^{-1} = (ghg^{-1})N \in M/N$.
Therefore, $M/N$ is normal in $G/N$.

The third isomorphism theorem states that for $N \le M \le G$ with $N$ and $M$ being normal subgroups of $G$, then:


*

*$M/N$ is normal in $G/N$.

*$(G/N)/(M/N) \cong G/M$


So in fact this is part of the third isomorphism theorem itself, not its converse.
