Let $G$ be a finite abelian group of odd order. Which of the following define an automorphism of $G$?
a. The map $x→ x^{−1}$ for all x ∈ G.
b. The map $x→ x^2$ for all x ∈ G.
c. The map $x→ x^{−2}$ for all x ∈ G.
I have verified that all of them are homomorphism.
(a) it is bijective since each element in a group must have inverse and it is unique.
But I could not verify other two are bijective or not.
how can I able to solve this?
$2$.let $g\in G$ be an arbitrary element and $o(g)=l$. So there are $x,y\in\mathbb Z$ such that $2x+ly=1$. Thus $g=g^{2x}$.
$3$.As you know, the composition of two automorphisms are an automorphism.
For part (b), let $x^2=y^2$ for some $x,y\in G$. Then $(xy^{-1})^2=e$. If $x\neq y$, then $xy^{-1}\in G$ has order 2, which is not possible. So we must have $x=y$. Thus $x\rightarrow x^2$ is one-one. Now, using the fact that any self-map on a finite set is injective if and only if it is surjective, it follows that $x\rightarrow x^2$ is an automorphism. The map in part (c) is a composition of the automorphisms defined in part (a) and (b), and hence an automorphism.
The kernel of both your homomorphisms consists of all the elements whose square is one. Any such element, other than one, is the generator of a cyclic subgroup of order two. But according to Lagrange, there can be no such subgroup, so the kernel is trivial, and the homomorphisms are automorphisms.
