1. ## Group Problem

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

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

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

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

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

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