The converse of Factor Group criterion

Question

If N is a normal subgroup of G then G/N is a group. Is the converse true? I mean If G/N is a group then N is Normal.

Answer 1

The statement is true. As if you will take canonical homomorphism $\eta :G \to G/N$ , Then $ker \eta=N$. And $ker$ is always a normal subgroup.

$\textbf{Without homomorphism-}$ We now just use the definition of normality. Let $g \in G$, then all we need to show is $gNg^{-1} \in N$. So consider these $3$ elements of $G/N$, namely $gN$, $xN$ and $g^{-1}N$, where $x \in N$ and then multiply them all, we get $(gxg^{-1})N=(gN)(xN)(g^{-1}N)=(gN)(N)(g^{-1}N)=(gg^{-1})N=N $ implies $gxg^{-1} \in N$, and thus $N $ is normal in $G$. [above I used $xN=N,$ as $x \in N$ as you must know why this. Right?]