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

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

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

• Doesn't first isomorphism theorem do the trick under the mapping $(m, n) \mapsto m$? Dec 24, 2018 at 4:24
• What about R - the normal subgroup? Dec 24, 2018 at 4:25
• Right so FIT states if we have a group homomorphism $\phi$ from $M \to N$, then: $M / \ker(\phi) \cong Im(N)$. That gives us exactly what we want, where $\ker(\phi)$ gives us $R$. Dec 24, 2018 at 4:28
• Also I hastily ended up answering this, but please provide what you tried next time. Dec 24, 2018 at 4:30

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