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,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
{1,a,a2,a3,b,a2b,c,a2c}
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

Log in or registerto write something here or to contact authors.