# Homomorphism 6

• April 24th 2009, 07:51 AM
mpryal
Homomorphism 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.)
• April 24th 2009, 10:23 AM
ThePerfectHacker
Quote:

Originally Posted by mpryal
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.)

Let $\pi : G\to G/N$ be the natural projection homomorphism i.e. $g\mapsto gN$.
Let $a\in G$ then order of $\pi (a)$ must divide order of $a$ by the properties of homomorphism.
Thus, we see that $m | \text{o}(a)$.