Prove that every group of order $4$ is abelian as follows.

Another strategy to 1)

If there exist some $a \in G$ such that $O(a) = 4$ then $G = \{e,a,a^{2},a^{3} \}$

Thus $c,d \in G \implies c= a^{i}, d = a^{j} \implies c\cdot d = a^{i} \cdot a^{j} = a^{i+j} = a^{j} \cdot a^{i} = d\cdot c$

3) Suppose there no exist $a \in G$ such that $O(a) = 4$

Then if $a \in G, a \not= e \implies O(a) = 2$

Let $a,b \in G$ then $a \cdot b \in G \implies (a\cdot b)^{2} = e \implies (a\cdot b)(a\cdot b) = e \implies a\cdot b \cdot a \cdot b = e $

Thus $a \cdot (a\cdot b \cdot a \cdot b) \cdot b = a \cdot e \cdot b = a \cdot b \implies a^{2}\cdot b \cdot a \cdot b^{2} = a\cdot b \implies e \cdot b \cdot a \cdot e = a \cdot b$

$ \implies b \cdot a = a \cdot b $

Then $G$ is an abelian group.