### Finally, a valid generalisation of the Third Isomorphism Theorem

Let $G$ be a commutative group. Let $M$ and $N$ be subgroups of $G$. If $M\leq N$, then $\frac{(G/M)}{(N/M)}\cong \frac{G}{N}$.

However, what if $M\not\leq N$? We will consider the general scenario, where $M$ and $N$ are any subgroups of $G$, provided $N$ is not a proper subgroup of $M$.

Then $\frac{(G/M)}{(N/M)}\cong \frac{G}{N+M}$

In the case that $M\leq N$, we have $M+N=N$.

I have arrived upon this result myself. I don’t know if it is already known.