The

groups of order 8 are split into

Let's sketch the proof that these last two are the only
nonabelian ones.

Let *G* be a group of order 8. By Lagrange's Theorem the order of an element
of *G* divides the order of *G*, in this case 8,
so each element has order 1,2,4 or 8.
If there is an element of order 8 then *G* is cyclic.
On the other hand if there is no element of order 4 then *G*
must consist of elements with order 2 (and the identity). It is
easy to see that such a group is abelian.

So for *G* to be nonabelian it must have an element of order
4, call it *a*. This gives us 4 elements of *G*, viz
*1,a,a*^{2},a^{3}. If there is an element
different from one of these with order 2, *s* say, then
since *sas*^{-1} also has order 4 and
*<a>* is a normal subgroup (having only 2 cosets).
This forces that *sas*^{-1}=a^{-1}.
This makes it clear that *G* is isomorphic to the Dihedral group *D*_{4}.

So this leaves us with the case that all the elements of *G*
different from the powers of *a* must have order 4.
Let *b* be one of these. Consider *b*^{2}.
This has order 2 and so it must equal *a*^{2}

Now let *c=ab*. Note that *c* is not a power of *a*
or *b*. So again *c* has order 4.

The elements of *G* are therefore

*{1,a,a*^{2},a^{3},b,a^{2}b,c,a^{2}c}

At this point it is clear that this group is isomorphic
to the Quaternion group. The isomorphism is given by
mapping *i,j,k* to *a,b,c* and -1 to *a*^{2}