Show that $(M\times N)/R\cong M$.

Question

If there are two groups, $M$ (with multiplication $\cdot_M$) and $N$ (with multiplication $\cdot_N$) and we define a new group $M \times N$ with multiplication such that $$ (m,n)(m',n') = (m \cdot_M m',n \cdot_N n') $$ There can be a normal subgroup $R$ of $M \times N$ such that

• $eR = \text{identity element}$

• $R = \{(eR, r) | r \in R\}$

Show that

$$ \frac{M \times N}{R} \simeq M $$

Answer 1

Use the first isomorphism theorem under the mapping $\phi: M \times N \to M$ that takes $(m,n) \mapsto m$. This map is surjective onto $M$, and has kernel $R$ (since whenever $n = e$, we get mapped to the identity). Thus:

$$ (M \times N) / \ker(\phi) \cong Im(\phi) \implies (M \times N)/R \cong M $$

You can check it to be a group homomorphism.