Let $G$ is a group and $H \leq G$ and $x \in G$
Show that number of $G$'s right residue classes are equal to left ones wrt $H$
1) There is a hint in my notes " use $f: Hx \to x^{-1}H$ but using $f : Hx \to xH$ looks better for me. Is it wrong?
If it is not, should I put elements in function like below?
2) For instant, should I write $" f(h_1x)=f(h_2x) \Rightarrow h_1=h_2$ when showing being (1-1)?
3) Last, how can I show surjectivity in the best way? We have never explain surjectivity in detail in our examples.
In addition there is a little exercise.
Take $G=S_4$ and $H=\{I,(12)(34),(13)(24),(14)(23)\}$ where $I$ is identity function. Find all right residue classes wrt $H$
4) We know by Lagrange's Theorem that the number of right residue classes is $6$. Do I have to multiply all elements one by one or is there anything that I can do quickly?
Any help'd be very beneficial for me. Thanks in advance
The reason why you would want to use $f: Hx \rightarrow x^{-1}H$ is because you can show that it is simply a bijection where:
$$f(Hx) = \{ (hx)^{-1} : h \in H \} = \{ x^{-1}h^{-1} : h \in H \} = x^{-1}H $$
So it's similar to yours, except in your approach there is no clear way to identify the left and right actions on $H$. Instead, I simply use the fact that $x$ has an inverse that we can immediately use.