Originally Posted by

**skyking** Heres the problem:

let $\displaystyle G$ be a non-abelian group of order 8

1. prove there exists an element of $\displaystyle G$, call it $\displaystyle a$ such that it's order is 4.

2. Let $\displaystyle b$ be an element of $\displaystyle G$ such that $\displaystyle b$ is not formed by $\displaystyle a$. Show that $\displaystyle a^{-1}=a^3=b^{-1}ab$.

3. Show that if the order of $\displaystyle b$ is 2 then $\displaystyle G$ is isomorphic to $\displaystyle D_{4}$, and if the order of $\displaystyle b$ is 4 then $\displaystyle G$ is isomorphic to the quatrernion group.

My solution so far:

I was able to show 1 easily by using legrange's theorem and eliminating the possiblities of 8,1,2 being the only order of elements.

Since $\displaystyle a$ is of order 4 it is obvious why $\displaystyle a^{-1}=a^3$.

To show 2 I showed that the elements of $\displaystyle G$ must be $\displaystyle e,a,a^2,a^3,b,ab,a^2b,a^3b$. Then my idea is to show by elimination that $\displaystyle ba^3=ab$.

I was able to eliminate all options besides $\displaystyle ba^3=a^2b$ and this is where I'm stuck.

$\displaystyle ba^3=a^2b\Longrightarrow a^3=b^{-1}a^2b\Longrightarrow a^2=(a^3)^2=b^{-1}a^2bb^{-1}a^2b=b^{-1}a^4b=1\Longrightarrow $ contradiction.

Tonio

I'm not sure if this is the right approach at all or is there a much simpler and elegant way of showing this.

Thanks,

SK