1
$\begingroup$

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

  • 1
    The subgroup being normal is not relevant here. – 2017-01-15
  • 0
    @TobiasKildetoft:You meant to say that the above result is also valid if $H$ is any arbitrary subgroup of $G$? – 2017-01-15
  • 2
    Yes, except we usually only use the notation $o(\cdot)$ and call it the order when it is a group (and even then most authors use $|\cdot|$ in all cases as it is more readable). – 2017-01-15
  • 0
    @TobiasKildetoft:Still i'm not getting how to proceed. – 2017-01-15
  • 1
    It's the well-known Lagrange Theorem for Group Order. A proof of it can be found in any group theory book. A quick search on the Internet would do it too. – 2017-01-15
  • 0
    Yes, [Lagrange](https://en.wikipedia.org/wiki/Lagrange%27s_theorem_(group_theory)) is your friend. – 2017-01-15

2 Answers 2

0

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

0

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 $|G|/|H|=r=|G/H|$