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Finite groups up to order 9

To list all the finite group up to order 9, we use the following results:

Theorem. A finite group of a prime number order is cyclic.

Consequently there is only one group of each of the orders , namely the cyclic groups ,, and .

The groups of order 4 and 9 are covered by the theorem:

Theorem. Let the order of a group is , where is a prime number. Then is Abelian. Moreover either is cyclic of order , or it is a direct product of two cyclic groups of order .

This result implies that there are two groups of order 4 and 9. These are groups cyclic groups or , or , and . Each of them is Abelian.

The group is also known as the Klein four group .

Let be a group of order 8. Lagrange's Theorem implies that  the order of an element of divides the order of , that is 8. Therefore each element of has order 1,2,4 or 8.

If there is an element of order 8 then is cyclic, i.e. .

If contains no element of order 4 then must consist of elements with order 2 (and the identity). It is easy to see that such a group is abelian. Namely, in this case , i.e. for every . Then for every .

So for to be non-Abelian it must have an element of order 4, say . This gives us 4 elements of : . If there is an element different from one of these with order 2, say, then since also has order 4 and the group generated by is a normal subgroup (since it has only 2 cosets). This implies:

• is the left coset decomposition of and . This implies that and that .
• . Since od of order 4, either or . The first case implies , and that or that is Abelian, a contradiction.  

Consequently . Since and , the equalities and are impossible. In both cases , and since , would be Abelian. Therefore it remains that and and the resulting group is the Dihedral group .

The group is generated by two elements and , where is of order 4, of order 2, such that .

So this leaves us with the case that all the elements of not belonging to must have order 4. Let be one of these. Since has order 2, it must equal . The element cannot be a power of or . So again has order 4. The elements of are therefore . This group is isomorphic to the Quaternion group .

The group is generated by two elements and , both of order 4 such that and . When using quaternion unities to describe , the isomorphism is given by mapping to and to .

For groups of order 6 we have two possibilities: the cyclic group or the direct product of two cyclic groups .

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