= 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 $1,r,r^2,r^3,s,sr,sr^2,sr^3$. Two generators are minimal because this group is not cyclic. Its subgroups of order two are $\langle r^2\rangle$ and $\langle sr^j\rangle$ for $j=0,1,2,3$. Its subgroups of order four are $\langle r\rangle$, $\langle r^2,s\rangle$ and $\langle r^2,sr\rangle$.
A <group homomorphism> obeys $\phi(gh)=\phi(g)\phi(h)$ for all elements; a <group isomorphism> is a bijective group homomorphism. Take the rotation subgroup $\langle r\rangle$ and define
$$
\boxed{\phi(r^j)=i^j\in Q,\qquad \ker\phi=\{1\}.}
$$
The 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 \b[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 $G=\langle g\rangle$ 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 $\langle g^d\rangle$ has order $n/d$, strictly between one and $n$. This proves the assertion described by <groups with no proper nontrivial subgroups>.
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