Let G be a finite abelian group. I am asked to find sufficient conditions for
$$ \begin{array}{cccc}
\alpha: & G & \rightarrow & G & \\
& g & \mapsto & g^{2}
\end{array} $$
to be a group isomorphism.
I already know that if the order of the group is odd, then $ \alpha $ is injective, and thus bijective.
Is there any way to prove this without using cyclic groups and Lagrange's Theorem?
I am supposed to use just basic things like definitions and equivalence relations.
Best Answer
One way to prove that $g\mapsto g^2$ is not an isomorphism for even cardinality finite groups directly from the group axioms is to consider the map $x\mapsto x^{-1}$. This has order $2$, and if $|G|$ is even, must have an even number of fixed points, since points are either fixed or in a pair, since this map is order $2$. But we know the identity is a fixed element, so there must be a non-identity element fixed by this, that is, some nontrivial $g\in G$ such that $g^{-1}=g$. For this $g$, we have $g^2=e=e^2$, so the map can't be injective if the order of $G$ is even. Note, we didn't need $G$ to be abelian for this argument.