If G is a group and N is a normal subgroup of G, show that if a in G has finite order o(a), then Na in G/N has finite order m, where m|o(a). (Prove this by using the homomorphism of G onto G/N.)

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- Apr 24th 2009, 06:51 AMmpryalHomomorphism 6
If G is a group and N is a normal subgroup of G, show that if a in G has finite order o(a), then Na in G/N has finite order m, where m|o(a). (Prove this by using the homomorphism of G onto G/N.)

- Apr 24th 2009, 09:23 AMThePerfectHacker
Let $\displaystyle \pi : G\to G/N$ be the natural projection homomorphism i.e. $\displaystyle g\mapsto gN$.

Let $\displaystyle a\in G$ then order of $\displaystyle \pi (a)$ must divide order of $\displaystyle a$ by the properties of homomorphism.

Thus, we see that $\displaystyle m | \text{o}(a)$.