Problem: Let G be a non-Abelian group with 8 elements
– Show that G has an element $a$ of say, order 4.
– Let $b$ be an element of G that is not $e$, $a$, $a^2$ or $a^3$. By considering the possible values of $b^2$ and of $ba$ and of $ab$ show that G is isomorphic either to the dihedral group or to the quaternion group.
Approach: I already proved the first thing by using Lagrange's theorem and showing that if there were no elements of order 4, then all elements would have order 2 $\Rightarrow$ (G is Abelian) which is a contradiction. For the second part, I am not entirely sure how to proceed, but I know that the only 2 non-abelian groups of 8 elements are both D(4) (dihedral group) and $\mathbb{H}_0$ (quaternion group).
Best Answer
First, note that $$\{e,a,a^2,a^3,b,ab,a^2b,a^3b\}$$ are all distinct. Also $$b^2\in\{e,a,a^2,a^3\}$$ Suppose $b^2=a $ or $a^3$. Then $b$ will be of order $8$. Then $G$ is cyclic; a contradiction. Thus $$b^2\in \{e,a^2\}$$ Next,$$ba\in\{ab,a^2b,a^3b\}$$ Suppose $ba=a^2b$. Then $bab^{-1}=a^2\implies ba^2b^{-1}=1\implies a^2=1$, a contradiction. Also, since $G$ is not abelian $ba\neq ab$. Thus $$ba=a^3b$$ So there are two combinations here. $b^2=a^2$ and $ba=a^3b$ will give $D_8$ while $b^2=e$ and $ba=a^3b$ will give $Q_8$