Every Abelian group G, of order 6, is cyclic.

“Cyclic” just means there is an element of order 6, say a, so that G={e,a,a²,a³,a⁴,a⁵}. More generally a cyclic group is one in which there is at least one element such that all elements in the group are powers of that element.

Proof:

The order of each non-identity element is 2, 3, or 6. If we can show an element of order 6 we're done.

Part 1. There is at most 1 element of order 2. Suppose instead that there exist 2 elements of G, a and b, which are distinct elements of order 2 so that a²=e and b²=e Then ab≠a and ab≠b and ab≠e. WHY? But (ab)²=abab=aabb=e. Thus o(ab)=2. Now we can show that {1,a,b,ab} form a group. But that's a contradiction. WHY? Thus we've shown that there is at most one element of order 2 in an abelian group of order 6.

Part 2. If there is an element of order 6, the group is cyclic, so let's assume there are no elements of order 6. The five non-identity elements are of order 2 or 3. But we've shown that at most one is of order two. Hence there is at least one element of order 3. But the elements of order 3 occur in pairs (WHY?); hence there are either 2 or 4 elements of order 3. Thus there is at least one element of order 2.(WHY?)

Part 3. From parts 1 and 2, we conclude that there is an element, say a, with o(a)=2 and an element, say b, with o(b)=3. Now we consider the element ab. What is the order of ab?

· ab≠e (WHY?).

· (ab)² ≠e since:(ab)²=abab=aabb=eb²=b²≠e)

· (ab)³ ≠e WHY?

· (ab)⁴≠e else b=e and that's a contradiction

· (ab)⁵≠e else {e,ab,(ab)² ,(ab)³ ,(ab)⁴} form a subgroup of G of order 5 and that's a contradiction. (There are other arguments why (ab)⁵≠e, try to find a proof of this that does not use Lagrange's Theorem.)

But o(ab)≤6. WHY?

Therefore o(ab)=6. Therefore G is a cyclic group of order 6. Later we'll show that all cyclic groups order 6 are essentially the same. So we'll have then shown that: There is essentially one abelian group of order 6.