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$$ \;$ $\forall \; g \in G$

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

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