If $G$ is a finite group and $H$ is a normal subgroup of $G$,then prove that $o(G/H)=o(G)/o(H).$

Question

If $G$ is a finite group and $H$ is a normal subgroup of $G$,then prove that $o(G/H)=o(G)/o(H).$

Since $G$ is finite, then $G=\{a_1,a_2,a_3,...,a_n\}$. Also, $H\lhd G\implies Ha_i=a_iH$ ,$i$ ranges from $1$ to $n$. Now, $G/H=\{a_1H,a_2H,a_3H,...,a_nH\}$.

I got stuck here.I'm not getting how to use this information in order to get the required result.

Any hints are heartly welcome.

Thank you

Answer 1

You don't need $H$ to be normal in $G$. Just use the fact that there is a bijection between any two cosets, and that they form a partition on $G$. You can find this in any standard book on abstract algebra. See for instance PROPOSITION 1.25 in http://www.jmilne.org/math/CourseNotes/GT310.pdf

Answer 2

Let be $|G/H|=r$, We know that $G=\cup_{i=1}^{r}a_iH=a_1H\cup a_2H \cup...\cup a_rH $ then $|G|=\sum_{i=1}^{r}|a_iH|=\sum_{i=1}^{r}|H|=r|H|; |a_iH|=|H|$ then $|G|/|H|=r=|G/H|$