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...
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 $M/N$.