= Solution
\b[No.] The <quaternion group>
$$
Q_8=\{1,-1,i,-i,j,-j,k,-k\},\qquad
i^2=j^2=k^2=ijk=-1,
$$
is another <non-abelian group> of order eight: for example $ij=k$ while $ji=-k$. Its multiplication comes from the associative <quaternion> algebra, so it does define a <group>.
It is not <isomorphic> to the <dihedral group> of the square. In $Q_8$, only $-1$ has order two; the six elements $\pm i,\pm j,\pm k$ have order four. The square's <dihedral group> has five elements of order two: four reflections and the half-turn. <Group isomorphisms> preserve element orders, so these groups are distinct. The hypothesis in part (c) excludes $Q_8$, since its only element of order two is already in each cyclic subgroup of order four.
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