Show that $\alpha: G \rightarrow G ,\alpha(g)=g^2$ isomorphism

Question

Let $G$ be a group and $\alpha: G \rightarrow G ,\alpha(g)=g^2$ ,$G$ is abelian group, $|G|=2n+1$ show that $\alpha$ isomorphism

Attempt:

$\alpha (g_1g_2)=(g_1g_2)(g_1g_2)=(g_1g_1)(g_2g_2)$

$\alpha(g_1)\alpha(g_2)=(g_1g_1)(g_2g_2)$

$G$ is abelian $\Rightarrow \alpha$ is homomorphism

Answer 1

Since you have proved that $\alpha$ is an homomorphism, and $|G|$ is finite, it suffices to show that $\alpha$ is 1-to-1, or equivalently, that $Ker(\alpha)=\{1\}$.

But if $\alpha(g)=g^2=1$, and $g\neq 1$ then the order of $g$ is $2$, that is, $\{1,g\}$ is a subgroup, and this is impossible because $2$ does not divide $|G|=2n+1$.

Answer 2

You may want to say why $G$ abelian implies $\alpha$ is a homomorphism. It's not too hard to show that $\alpha$ is an isomorphism. Since, $G$ is finite, it suffices to show that $\alpha$ is injective.

Let $g\in\ker\alpha$ ($e$ the identity element)

$$\begin{align} g\in\ker\alpha & \implies g^2=e\\ & \implies g=e\text{ or } |\langle g\rangle|=2. \end{align}$$

If $g$ has order $2$, $G$ has a subgroup of order $2$ which is impossible because the order of $G$ is odd. So, $\ker\alpha=\{e\}$.

Answer 3

Hint: Use Lagrange's theorem to deduce that $G$ has no element of order $2$.