The groups of order 8
are split into
and the nonabelian groups of order 8.
Let's sketch the proof that these last two are the only
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,a2,a3. 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 D4.
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 b2.
This has order 2 and so it must equal a2
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
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 a2