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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).

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  • $\begingroup$ Let $C(a)$ be the cyclic group generated by $a$. This group, therefore, has order $4$, hence has index $2$, hence is a normal subgroup of $G$. The behavior of $G$ is now completely determined by how $b$ acts on $C(g)$ by conjugation; namely, does this action have 1 orbit or 2? $\endgroup$ – avs Nov 10 '16 at 0:37
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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 $Q_8$

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  • $\begingroup$ I think you meant $b^2 = a^2$ will give $Q_8$ $\endgroup$ – Gabriel B. H. Lisboa Sep 26 '18 at 4:55
  • $\begingroup$ @GabrielB.H.Lisboa corrected. $\endgroup$ – Alan Wang Sep 27 '18 at 5:09

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