Proof: For each $a$, the function $\gamma_a$ is a permutation of the set $G$. For $a\in G$ we can define a function $\gamma_a:G\to G$ given by $\gamma_a(x)=axa^{−1}$ for $x\in G$.
Then for each $a$, the function $\gamma_a$ is a permutation of the set $G$.
Proof:
By definition, a permutation of a set, say $X$, is a bijective function from $X$ to itself.
So, since $\gamma_a$ is defined from to $G$ to $G$ we only have to show that it is bijective.
Surjection: Let $y\in G$ and since the function maps elements in $G$ onto $G$ we have $axa^{−1}=y$. This implies $x=a^{−1}ya$.
So, we have $$\gamma_a(a^{−1}ya)=a(a^{−1}ya)a^{−1}=aa^{−1}yaa^{−1}=y.$$
Injection: For any $x,y\in G$ we have $$\gamma_a(x)=\gamma_a(y).$$ This implies $axa^{−1}=aya^{−1}$ and using the cancellation laws we get $x=y$.
Therefore, $\gamma_a$ is a bijective and so, a permutation of the set $G$.
My question is if my arguments above are correct. I would appreciate any help.
Thank you.
 A: You're correct, but some of your wording is slightly odd and makes the argument harder to read than it should be (I don't know if you're a native English speaker, which might be the problem). For example:

Injection: for any $x, y \in G$ we have…

This means "it is true that for any $x, y \in G$", when what you mean is the following:

Injection: Suppose we had $x, y \in G$ with $\gamma_a(x) = \gamma_a(y)$…

Additionally:

Surjection: Let $y \in G$ and since the function maps elements in $G$ onto $G$ we have $axa^{−1}=y$. This implies $x=a^{−1}ya$.

You mean something like "Let $y \in G$, and set $x = a^{-1} y a$. Then $\gamma_a(x) = \gamma_a(a^{-1} y a) = y$. Therefore, since $\gamma_a$ hits every $y \in G$, it is surjective." The way you've written it, you've basically assumed your conclusion in saying $axa^{-1} = y$.
(An instance which does not affect readability: you've missed out the word "function" in "Therefore, $\gamma_a$ is a bijective  and so, a permutation of the set $G$." Alternatively, you could use "$\gamma_a$ is a bijection".)
