Prove the function $F: G \to{\rm Sym}(G)$ given by $F(g) = R_{g^{-1}}$ is an injective homomorphism The problem statement is as follows:

Let $G$ be a group and for all $g \in G$, let $R_g: G \to G$ be right multiplication given by $$R_g(h) = hg$$ Prove that the function $F: G \to{\rm Sym}(G)$ given by $$F(g) = R_{g^{-1}}$$ defines an injective homomorphism.

My main qualm with this question is through the course of proving homomorphism I seem to go from proving the homomorphism quality of the Right multiplication in general to needing them to act on an element of $G$ which in this case I call $f$ in order to prove the property. Is there some way to show this without relying on having right multiplication act on an element? I feel like I'm stepping outside the bounds of the function $F$ here.
For instance, it was my understanding that when proving the homomorphism property we want to prove it for the elements in the domain so in this case we would want to show: $$F(gh) = F(g)F(h),\ \forall g,h \in G$$ Correct? Now for this:
$$\begin{align}
F(gh) &= R_{g^{-1}h^{-1}} \\
&= R_{g^{-1}h^{-1}}(f)\\
& = fg^{-1}h^{-1} \\
&= (fg^{-1})h^{-1} \\
&= (R_{g^{-1}}(f))h^{-1} \\
&= R_{h^{-1}}(R_{g^{-1}}(f)) \\
&= R_{h^{-1}} \circ R_{g^{-1}}(f)\\
& = R_{h^{-1}} \circ R_{g^{-1}}
\end{align}$$
for all $g,h,f \in G$, which would then imply $F(gh) = F(g)F(h)$?
For injectivity it seems the simplest method to show a homomorphism is injective is to show that the kernel is trivial. Since the kernel is all elements from the domain which are sent to the identity in the image then we would want to find all elements which are sent to the identity permutation $\text{id}_{{\rm Sym}(G)}$ so in ${\rm Sym}(G)$ we have the identity is $$R_g(h) = hg = h.$$ Hence $\text{id}_{{\rm Sym}(G)} = R_e = F(e)$ since the inverse of the identity in a group is the identity itself.
Correct?
 A: You should be more careful about showing $F(gh)=F(g)F(h)$. For one, you're working with maps: $F(g)$ is a map, while $hg^{-1}$ is not: it's an element of $G$, while $F(g)$ is an element of $\operatorname{Sym}(G)$. You should not connect them with an equals sign, because they are not equal. You have to work with at least two chains of equations. First: You want to show that
$$F(gh)=R_{(gh)^{-1}}=R_{g^{-1}}\circ R_{h^{-1}}=F(g)F(h).$$
It's the middle equality that's the crux. It's an equality of maps, and two maps are equal if they map every element $f$ to the same image. So you need to show that $R_{(gh)^{-1}}(f)=R_{g^{-1}}(R_{h^{-1}}(f))$ for all $f$ in $G$. This is an equality of elements of $G$, and should not be inserted into the chain of equalities above!
Anyway, here you made you second mistake: $(gh)^{-1}=h^{-1}g^{-1}\neq g^{-1}h^{-1}$. So we have
$$\begin{align}R_{(gh)^{-1}}(f)&=fh^{-1}g^{-1}\\
&=(fh^{-1})g^{-1}\\
&=R_{g^{-1}}(fh^{-1})\\
&=R_{g^{-1}}(R_{h^{-1}}(f))\\
&=R_{g^{-1}}\circ R_{h^{-1}}(f)
\end{align}$$
for all $f\in g$, making the maps $R_{(gh)^{-1}}=R_{g^{-1}}\circ R_{h^{-1}}$ as desired.
Your proof for injectivity is fine.
A: Firstly, is $F(g)$ indeed a bijection on $G$, for every $g\in G$ ($F$'s good definition)? Yes, because:

*

*injectivity: $F(g)(h)=F(g)(h')\Longrightarrow hg^{-1}=h'g^{-1}\Longrightarrow h=h'$;

*surjectivity: for every $h'\in G$, there is $h\in G$ such that $h'=F(g)(h)$ if and only if $h'=hg^{-1}$ if and only if $h=h'g$.

Now, the homomorphism part. For every $g,g',h\in G$, we get: $F(gg')(h)=h(gg')^{-1}=$ $hg'^{-1}g^{-1}=$ $(hg'^{-1})g^{-1}=$ $F(g)(hg'^{-1})=$ $F(g)(F(g')(h))=$ $(F(g)F(g'))(h)$, whence $$F(gg')=F(g)F(g')$$ Finally, the injectivity of $F$ itself: $$F(g)=F(g')\iff\forall h\in G:F(g)(h)=F(g')(h)\iff\forall h\in G:hg^{-1}=hg'^{-1}\Longrightarrow g^{-1}=g'^{-1}\Longrightarrow g=g'$$
