Prove that a group is abelian under given conditions Let $(G, \cdot)$ be a group with the property that there exists $a \in G$ such that $ax^3a = x$, $\forall x \in G$. Prove that $(G, \cdot)$ is abelian.
Here is my solution:
By plugging in $x := ax$ in the given condition, we get $a^3x^3a = x$. But $ax^3a$ is also equal to $x$, so $a^3x^3a = ax^3a$ $\Rightarrow$ $a^2 = e$. That means that $a = a^{-1}$.
Now, using $a = a^{-1}$ in the given condition, that is $ax^3a = x$, we get $ax^3 = xa$. By squaring both sides, we get $a^2x^6 = x^2a^2$. But we know that $a^2 = e$, so $x^6 = x^2$, or, equaivalently, $x^4 = e$.
I know that if $x^2 = e$, $\forall x \in G$, then the group is abelian. But does this hold if $x^4 = e$?
Thank you!

There are a couple of issues with your argument, but it starts with

By plugging in $x := ax$ in the given condition, we get $\color{red}{a^3x^3a = x}$.

The equation in red is incorrect. If you replace $x$ with $ax$, then you would get $a(ax)^3a = ax$. 
Replace $x$ with $axa^{-1}$ in the condition to get $a(axa^{-1})^3 a = axa^{-1}$, or $a^2x^3a^{-2} = axa^{-1}$. Cancellation gives $ax^3a^{-1} = x$; in particular, setting $x = a$ yields $a^3 = a$, or $a^2 = e$. Now $$(yx)^3 = a^{-1}(yx)a = (a^{-1}ya)(a^{-1}xa) = y^3x^3,$$ that is, $y(xy)^2x = y^3x^3$. By cancellation, $(xy)^2 = y^2x^2$. Thus, 
$$x^2y^2 =  ax^6a^{-1} ay^6a^{-1} = ax^6y^6a^{-1} =a(y^2x^2)^3a^{-1} = y^2x^2.$$
Hence $(xy)^2 = x^2y^2$, i.e., $x(yx)y = x^2y^2$. Cancellation results in $yx = xy$. Consequently, $G$ is abelian.
