Can you please check my logic and maybe suggest some other strategies for the following problem:
Prove that every group of order $4$ is abelian as follows: Let $G$ be any group of order $ 4$, i.e., $|G| = 4$.
(1) Suppose there exists $a \in G$ such that $o(a) = 4$. Prove that $G$ is abelian.
(2) Suppose that no element of $G$ has order 4. Prove that $\forall x\in G$, $x^2 = 1$.
(3) Suppose that no element of $G$ has order 4. Prove that $G$ is abelian.
What I got so far:
(1) If there exists $a \in G$ such that $o(a) = 4$,
Case 1: $a\cdot a=b$. Then $a\cdot a\cdot a=c$ and $a\cdot a\cdot a\cdot a=1$. Algebra... G is abelian.
Case 2: $a\cdot a=c$... $G$ is abelian.
(2) Let $x\in G$. If $o(x) \neq 4$, we can clarify that an element cannot have an order greater than $4$ in a group of order $4$ and that the only element that has an order of $1$ is $1$. Therefore the other three elements must have an order of $2$, so $x^2=2$ for all $x \in G$.
(3) No ideas yet :(