I Quick question about Lagrange's theorem

[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:

This can be shown using the concept of left cosets of $H$ in $G$. The left cosets are the equivalence classes of a certain equivalence relation on $G$ and therefore form a partition of $G$. Specifically, $x$ and $y$ in $G$ are related if and only if there exists $h$ in $H$ such that $x = yh$. If we can show that all cosets of $H$ have the same number of elements, then each coset of $H$ has precisely $|H|$ elements. We are then done since the order of $H$ times the number of cosets is equal to the number of elements in $G$, thereby proving that the order of $H$ divides the order of $G$. Now, if $aH$ and $bH$ are two left cosets of $H$, we can define a map $f : aH \rightarrow bH$ by setting $f(x) = ba^{-1}x$. This map is bijective because its inverse is given by $f^{-1}(y)=ab^{-1}y$

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?

 

[fresh_42]

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?

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.

 

[TeethWhitener]

Cool, thanks.