I will try and show that if our group $G$ is not cyclic then it is isomorphic to Klein's four-group $V$.
So suppose $G$ is not cyclic. Then its elements, excluding $e$, are of order $3$ or $2$. Supposing there is an element of order $3$ gives three distinct elements $g, g^2, g^3$. By hypothesis there exists $h \in G$ such that $h$ is distinct from these three elements. So we can make $gh$ which must be equal to one of the elements previously mentioned. Setting this equal to the various elements:
\begin{align}
&gh=g \Rightarrow h=e && gh=g^2 \Rightarrow h=g & \\
&gh=h \Rightarrow g=e, && gh=g^3 \Rightarrow h=g^2 &
\end{align}
Each of which is a contradiction.
Therefore all the elements are of order two and we can get an isomorphism between $G$ and $H \times H$ where $ H$ is cyclic of order two. But this direct product is isomorphic to Klein's four-group so $G$ is isomorphic to $V$.
Is this correct?
Also it is just given in the book that $V$ is isomorphic to this product; how do I show this without checking that $f(gh)=f(g)f(h)$ for all combinations?
Best Answer
An idea: let $\,G=\{1,a,b,c\}\,$ be a non cyclic group of order $\,4\,$, so we can assume $\,a,b\,$ are not powers of each other.
The question is: what is $\,ab\,$?? You should find pretty easy to show this must be $\,c\,$ and, with a few lines more, you prove both that $\,G\,$ is abelian and in fact the same (i.e., isomorphic to) the Viergruppe $\,V\,$.
Another idea: you seem to feel pretty comfortable using isomorphisms, Klein group and stuff: why won't you use directly Lagrange's Theorem and avoid all that long proof about the impossibility of an element of order 3?