Let $(G, \cdot)$ be a group with the property that there exists $a \in G$ such that $x^3 = xax$, $\forall x \in G$. Prove that the given group is abelian.
Here is how I did it: $$x^3 = xax, \, \forall x \in G$$ $$y^3 = yay, \, \forall y \in G$$
By multiplying with $x^{-1}$ in both sides of the first equality, we get $x^2 = ax = xa$. Analogous, we get $y^2 = ay = ya$.
So, $ax = xa$ and $ay = ya$ $\Rightarrow$ $xaay = yaax$ $\Rightarrow$ $x^2y^2 = y^2x^2$.
Now, $x^3y^3 = xaxyay$ $\Rightarrow$ $(xy)^2xy = x^2xyy^2$. But $(xy)^2 = (yx)^2$, so $y^2x^2xy = x^3y^3$. By multiplying in the right side with $y^{-1}$, we get $y^2x^3 = x^3y^2$. By rewriting this as $(yx)^2x = x(xy)^2$, and by doing the substitution $(xy)^2 = (yx)^2 = t \in G$, we get $tx = xt$. So, our group is abelian.
Is this a correct solution?
Thank you!
Your solution is correct but an overkill.
The group seems to be a trivial one:- $$x^3 = xax, \, \forall x \in G$$ $$x^3x^{-1} = xaxx^{-1}, \, \forall x \in G$$ $$x^2 = xa, \, \forall x \in G$$ $$x^{-1}x^2 = x^{-1}xa, \, \forall x \in G$$ $$x = a, \, \forall x \in G$$
Hence there is only one elemnt in the group, $a$ which must be the identity. The group is trivially albelian.
The question was too easy. Did you make a mistake somewhere?