If $N$ and $M$ are normal subgroups of $G$, show that $NM/M$ is isomorphic to $N/N \cap M$.

Question

If $N$ and $M$ are normal subgroups of $G$, show that $NM/M$ is isomorphic to $N/N \cap M$ (Problem 5, I. H. Herstein, pag 87). Really I haven't idea in this problem. Can anyone help me? Thanks in advance.

Answer 1

Note: I've provided the skeleton of the argument. Try to fill in the detail of each step on your own. If you hover your mouse over a box, it will reveal more information.

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First construct a homomorphism $\varphi: N \to NM / M$.

$\varphi(n) = nM$, which is in $NM / M$ because $n = n1 \in NM$.

Check that it's an epimorphism (an onto function).

Take an arbitrary $nmM \in NM / M$. Since $m \in M$, we have the equality of cosets $nmM = nM$, and the latter is $\varphi(n)$.

Analyze the kernel of $\varphi$.

$$ n \in \ker \varphi \quad\Longleftrightarrow\quad nM = M \quad\Longleftrightarrow\quad n \in N \cap M $$

Therefore, the homomorphism $\varphi$ descends to the isomorphism $\bar\varphi: N / (N \cap M) \to nM / M$.

Answer 2

Any element in $NM/M$ is of form $nmM$ where $n\in N$ and $m\in M$. Since $mM=M$, $nmM=nM$. Thus we have each element in $NM/M$ is of form $nM$ (upto equivalence classes).

Observe that this already tells you the isomorphism since if $n\in(N\cap M)$, $nM=M$. So the distinct classes of $NM/M$ are the elements in $N$ "mod" $(N\cap M)$.

Putting it concretely, look at the map which sends $nM\rightarrow n.(N\cap M)$.