# Group of small order

• Aug 13th 2007, 04:55 PM
r7iris
Group of small order
Let G be a non-Abelian group with eight elements. Show that G has an element a say of order 4. Let b be an element of G which is not e, a, a^2 or a^3. By considering the possible values of b^2 and of ba and ab, show that G is isomorphic either to the dihedral group ot to the quaternion group.
• Aug 13th 2007, 07:24 PM
ThePerfectHacker
Quote:

Originally Posted by r7iris
Let G be a non-Abelian group with eight elements. Show that G has an element a say of order 4. Let b be an element of G which is not e, a, a^2 or a^3. By considering the possible values of b^2 and of ba and ab, show that G is isomorphic either to the dihedral group ot to the quaternion group.

Let us do a slightly tougher problem.

"Determine all groups of order 8 up to isomorphism".

Now, if $\displaystyle G$ is abelian the problem is extremely easy by the powerful fundamental theorem of finitely generated abelian groups. In which case it is either:
$\displaystyle \mathbb{Z}_8 \ \ \ \mathbb{Z}_2\times \mathbb{Z}_4 \ \ \ \mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$.

The hard part is doing the non-abelian case. By Lagrange's theorem the order of an element divides the order of the group, in which case $\displaystyle 2,4,8$ are the possible orders. Now it cannot be $\displaystyle 8$ for that will imply the group is cyclic and hence abelian. We show it cannot be have all elements of order 2. Because then $\displaystyle (ab)^2 = abab=1\implies ab=ba$ a contradiction. Let $\displaystyle a$ have order 4. And consider the subgroup $\displaystyle \left< a \right>$ of order 4. And let $\displaystyle b\not \in \left< a \right>$ which means the cosets $\displaystyle \left< a \right> \mbox{ and }b\left< a \right>$ contain all the elements of $\displaystyle G$. This shows that $\displaystyle a,b$ generate the group $\displaystyle G$. Since $\displaystyle (G:\left< a \right>)=2$ it means $\displaystyle \left< a \right> \triangleleft G$ and the factor group $\displaystyle G/\left<a\right>\simeq \mathbb{Z}_2$. The last sentence implies that $\displaystyle b^2 \in \left< a\right>$. It cannot be that $\displaystyle b^2 = a \mbox{ or }b^2=a^3$ for that will imply that $\displaystyle b$ has order 8, which cannot be since $\displaystyle G$ is non-abelian group. Thus, there are two possibilities: $\displaystyle b^2 = 1 \mbox{ or }b^2 = a$. We mentioned that $\displaystyle \left< a \right> \triangleleft G$ so $\displaystyle bab^{-1}\in \left< a\right>$. So, $\displaystyle bab^{-1}=1\mbox{ or }a\mbox{ or }a^2\mbox{ or }a^3$. If $\displaystyle bab^{-1}=1\implies a=1$ a contradiction. If $\displaystyle bab^{-1}=a^2 \implies ba^2b^{-1}=1\implies a^2 = 1$ a contradiction. If $\displaystyle bab^{-1}=a\implies ab=ba$ a contradiction. Thus, $\displaystyle bab^{-1}=a^3\implies ba=a^3b$. All this work leads to two group presentations:
$\displaystyle (a,b : a^4 = 1, b^2 = 1, ba=a^3b)$
$\displaystyle (a,b : a^4=1 , b^2 = a^2, ba=a^3b)$