# I Quick question about Lagrange's theorem

1. Sep 22, 2016

### TeethWhitener

I was looking at the proof of Lagrange's theorem (that the order of a group $G$ is a multiple of the order of any given subgroup $H$) in Wikipedia:

I understand this proof fine, but I was wondering, instead of finding a bijection between cosets, is it enough to find a bijection between an arbitrary coset $gH$ and the subgroup $H$? So, for instance, we have a map $f: gH \rightarrow H$, where $f(x) = g^{-1}x$. The map is bijective, with inverse $f^{-1}(y) = gy$. Is there anything wrong with this?

2. Sep 22, 2016

### Staff: Mentor

No. Whether you show $|aH|=|bH|$ for arbitrary $a\, , \, b$ or $|aH|=|H|=|eH|$ for all $a$ doesn't make a difference. And $"="$ is transitive. It's simply a matter of taste.

3. Sep 22, 2016

### TeethWhitener

Cool, thanks.