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

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

  • $\begingroup$ Doesn't first isomorphism theorem do the trick under the mapping $(m, n) \mapsto m$? $\endgroup$
    – TrostAft
    Dec 24, 2018 at 4:24
  • $\begingroup$ What about R - the normal subgroup? $\endgroup$
    – user628956
    Dec 24, 2018 at 4:25
  • $\begingroup$ 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 $R$. $\endgroup$
    – TrostAft
    Dec 24, 2018 at 4:28
  • $\begingroup$ Also I hastily ended up answering this, but please provide what you tried next time. $\endgroup$
    – TrostAft
    Dec 24, 2018 at 4:30

1 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.


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