 Solution

Let $r$ be a quarter-turn and $s$ a reflection of the square. The dihedral group is generated by $r,s$ with $r^4=s^2=1$ and $srs=r^{-1}$, and its elements are $1,r,r^2,r^3,s,sr,s{r}^2,s{r}^3$. Two generators are minimal because this group is not cyclic. Its subgroups of order two are $⟨r^2⟩$ and $⟨s{r}^j⟩$ for $j=0,1,2,3$. Its subgroups of order four are $⟨r⟩$, $⟨r^2,s⟩$ and $⟨r^2,sr⟩$.

A group homomorphism obeys $\varphi (gh)=\varphi (g)\varphi (h)$ for all elements; a group isomorphism is a bijective group homomorphism. Take the rotation subgroup $⟨r⟩$ and defineThe quaternion element $i$ has order four, so the map is well defined modulo four, preserves products and is injective onto $\{1,i,-1,-i\}$. Surjectivity onto all eight quaternion elements is not required of a homomorphism.

A cyclic group is generated by the powers of one element. The final claim needs the usual qualification no nontrivial proper subgroup, and $|G|>1$: the trivial subgroup is proper in every nontrivial group, even one of prime order. With that interpretation, if $G=⟨g⟩$ has prime order $p$, Lagrange's theorem permits only subgroup orders one and $p$. Conversely, if its order $n>1$ is composite, choose a divisor $d$ with $1<d<n$. The subgroup $⟨g^d⟩$ has order $n/d$, strictly between one and $n$. This proves the assertion described by groups with no proper nontrivial subgroups.