Let $G=\{a_2, a_2, a_3...a_n\}$ be a finite abelian group of order $n$. Let $x=a_1 a_2 a_3...a_n$. Prove that $x^2=e$.
My solution: Well since the order of the group is finite, we know the order of the elements of the group are also finite. So there exists a certain $k \in \mathbb{Z}$ such that $a^k = e$. Since the order the $n$, it must be that $k = n$. So since $x=a_1 a_2 a_3...a_n$ then $x^2 =(a_1 a_2 a_3...a_n)^2$ and because the group is abelian we have $x^2=a_1^2 a_2^2 a_3^2...a_n^2$ but $a^n = e$ and $n=2$ (DID I GO WRONG HERE??) then we must have $x^2 = eee...e = e$
Let $G$ be a group and let $a \in G$. Prove $\langle a\rangle = \langle a^{-1}\rangle$.
I don't even know how to begin this one. I know the cyclic subgroup generated by $a$ is the set $\{a^n | n \in \mathbb{Z}\}$
Show that $U_5$ is isomorphic to $U_{10}$.
Well both of these have order 4. Theorem: Every finite cyclic group of order n is isomorphic to $\mathbb{Z}$. I don't know how to apply the theorem to the units.
General question: Are all unit groups ie, $U_5$ cyclic?