Let $G$ and $H$ be two groups of order 4 and neither group contains an element of order 4. Show that $G$ and $H$ are isomorphic. Let $G$ and $H$ be two groups of order 4 and assume further that neither group contains an element of order 4. Show that $G$ and $H$ are isomorphic.
Here is my attempt:
Let $G$ be a group order 4 which has no element of order 4. By Lagrange's theorem $o(g) \in $  {1,2,4}. Since there is no element of order 4, we have only 2 cases. If  $o(g) = 1$  , then $g = e$. On the other hand, if  $o(g) = 2$ ,  then $g \neq e$. So if $G$ equal to {a, b, c, d} then, $a^2 = b^2 = c^2 = d^2 = e$. Take $c = ab$. Since $c \neq a \neq b \neq e$ then, $a \leftrightarrow (1,0)$, $b \leftrightarrow (0,1)$ and $c \leftrightarrow (1,1)$.
Is my attempt correct? How can I finish the proof, and find the isomorphism? Any help would be great.

You still need to check that you have a homomorphism. That's what I kind of don't like about it.
I much prefer to use a sledgehammer in these situations.   I happen to know there are only two groups of order $4$.  They're both abelian.   The smallest non-abelian group is $S_3$.
Then no element of order $4$ means $G$ is not cyclic.   The only remaining possibility is $V_4$, the Klein four group.

I also like the hint.  It's easy to show that a group of order $p^2$ is abelian.   Then use the structure theorem.
