1. ## Group Problem

Let $G$ be a group. Suppose that
$(\alpha\ast\beta)^3=\alpha^3\ast\beta^3$ -----(1)
for all $\alpha, \beta \in G$.
(i)Show that for all $\alpha, \beta \in G$, then
$(\beta\ast\alpha)^2=\alpha^2\ast\beta^2$ and $\alpha^3\ast\beta^2=\beta^2\ast\alpha^3$.
(ii)If $G$ satisfies (1) and the condition that the identity element $e$ in $G$, is the only element in $G$ satisfying $x^3=e$, show that for all $\alpha, \beta \in G$,
$\alpha^2\ast\beta=\beta\ast\alpha^2$
and conclude that $G$ is abelian.

2. Proving part (i) is okay. The difficult portion is part (ii).

3. Originally Posted by Markeur
Let $G$ be a group. Suppose that
$(\alpha\ast\beta)^3=\alpha^3\ast\beta^3$ -----(1)
for all $\alpha, \beta \in G$.
(i)Show that for all $\alpha, \beta \in G$, then
$(\beta\ast\alpha)^2=\alpha^2\ast\beta^2$ and $\alpha^3\ast\beta^2=\beta^2\ast\alpha^3$.
(ii)If $G$ satisfies (1) and the condition that the identity element $e$ in $G$, is the only element in $G$ satisfying $x^3=e$, show that for all $\alpha, \beta \in G$,
$\alpha^2\ast\beta=\beta\ast\alpha^2$
and conclude that $G$ is abelian.
Hint 1. By condition (1), the map given by $f(\alpha) = \alpha^3$ is a homomorphism from the group to itself.

Hint 2. If the kernel of $f$ consists solely of the identity, then $f$ is injective. In particular, if $(\alpha^2\ast\beta)^3 = (\beta\ast\alpha^2)^3$ then $\alpha^2\ast\beta=\beta\ast\alpha^2$.

Hint 3. Use the properties in (i) to show that $(\alpha^2\ast\beta)^3 = (\beta\ast\alpha^2)^3$.

To conclude that G is abelian, notice that $(\alpha\ast\beta)^3 = \alpha^3\ast\beta^3 = \alpha^2\ast\beta^2\ast\alpha\ast\beta$, from which $(\alpha\ast\beta)^2 = \alpha^2\ast\beta^2$, and it easily follows that $\beta\ast\alpha = \alpha\ast\beta$.