Let $G$ be a nonabelian group of order 8 .
(a) Prove that $G$ must have an element of order $4,$ but none of order 8 .
(b) Let $a$ be an element of order $4,$ and let $N=\langle a\rangle .$ Show that there exists an element $b$ such that $G=N \cup N b$.
(c) Show that either $b^{2}=e$ or $b^{2}=a^{2}$. (Since $N$ is normal, consider the order of $N b$ in $G / N .)$
(d) Show that $b a b^{-1}$ has order 4 and must be equal to $a^{3}$.
(e) Conclude that either $G \cong D_{4}$ or else $G$ is determined by the equations $a^{4}=e$, $b a=a^{3} b, b^{2}=a^{2}$. Review Example 3.3 .7 (the quaternion group) to verify that the second case can occur.