# Let $N \unlhd G$ where $G$ is a finite group. Prove that order of $gN$ divides order of $g \; \forall \; g \in G$

Let $N \unlhd G$ where $G$ is a finite group. Prove that order of $gN$ divides order of $g$$\; \forall \; g \in G Define \phi : G \to G/N \phi(g) = gN \; \forall \; g \in G \phi is natural homomorphism Also by the property of homormorphisms if o(a) = n \Rightarrow o(\phi(a)) \mid n gN divides order of g$$ \;$ $\forall \; g \in G$

• is this proof fine ? – So Lo Dec 12 '17 at 8:01
• This is fine. Since you are learning these things, it is good for you to write everything out explicitly. – akech Dec 12 '17 at 8:14

Set the group $G' = G/N$.

Let $m$ be the order of $g$. Then $(gN)^{m} = g^{m}N = N = e_{G'}$.

It follows that the order of $gN$ divides $m$.