If $G$ is a finite group and $H$ is a subgroup of $G$, then $|H| = |gH|$ for every $g \in G$.

Question

I'm stuck. I believe I have half of the proof, but I'm missing an essential part to complete the proof. Any hints would be appreciated. Thanks!

Proof: Suppose $G$ is a finite group and $H$ is a subgroup of $G$. Let $H=\{ h_0,h_1,\ldots,h_n\}$. Now let $gH = \{gh_0,gh_1, \ldots gh_n\}$. Clearly $|gH| \le |H|$. Now to complete the proof, we take two elements $h_0, h_1 \in H$. Not sure what to use next...

Answer 1

Hint: define a map $\phi$ between $H$ and $gH$ by $\phi(h)=gh$. Show that this is a bijection.

Answer 2

HINT: Where you have $|gH|\ge|H|$, you should have $|gH|\le|H|$: there are clearly at most $|H|$ elements in the set $gH$, but it’s conceivable that $gh_i=gh_j$ for some $i,j\le n$ such that $i\ne j$. To finish, you need to show that this does not in fact happen. Suppose, then, that $i,j\le n$, and $gh_i=gh_j$. What happens if you multiply on the left by $g^{-1}$?

Answer 3

In response to the hints provided by Nicky Hekster and Brian M. Scott:

Assume, for $i$,$j$ $\le$ $n$, that $g$$h_i$ = $g$$h_j$. Then $g^{-1}$$g$$h_i$ = $g$$g$$h_j$, and $h_i$ = $h_j$. Thus |$H$| = |$g$$H$|.

Answer 4

Shorter (almost without any element): Note first the result makes sense for any part X of G, not merely subgroups.

As the map $\varphi\colon X \rightarrow gX$, $x \mapsto gx$, is onto, we have $\lvert gX\rvert \leq \lvert X\rvert$.

Apply this inequality to $gX$ and $g^{-1}$: $ \lvert g^{-1}(gX)\rvert=\lvert X\rvert\le\lvert gX\rvert$.