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).

  • $\begingroup$ 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

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 $Q_8$

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.