Show that every commutative group of order 6 is cyclic

Question

Q) Show that every commutative group of order 6 is cyclic.

I have no hint about where to begin in this question.

Since the group is commutative, we have $g_1g_2=g_2g_1$. How to proceed beyond this?

Answer 1

The outline of a proof is like that:

Assume there is no element of order 6. So all elements except e are of order 2 or 3.

a. Show elements of order 3 exist in pairs.

b. Show there is 4 element of order 3 and 1 element of order 2.

c. Let a have order 2 and b have order 3. Then consider ab. Show it has order 6.

Thus the group is cyclic.

Answer 2

You can use Cauchy's theorem or its proof.

By this theorem, there must be elements of order $2$ and elements of order $3$ in the group. Let $g_2$ and $g_3$ be elements of order $2$ and $3$, respectively.

Define $g=g_2g_3$.

Note that $g^k\neq e$, for $k=1,2,3,4,5$ and that $g^6=e$, the identity of the group.

Then $G=(g)$.

Answer 3

There are only two groups of order 6 less isomorphism (there are more groups, but the rest will be isomorphic to only two groups) $S_3$ and the $Z_6$. To reach this conclusion it is enough to employ Sylow's theorems. A group of order 6 will always have a normal subgroup of order 3 and index 2 in the group. In this way, the semidirect product of an order 3 cyclic or an order 2 cyclic is always defined. Evaluating the possible homomorphisms (there are only two possibilities) it is possible to conclude that the trivial homomorphism will characterize a direct product of an order 2 cyclic with one of order 3. As the order of the groupsa is coprima it follows that this group will be isomorphic to the cyclic group of order 6, therefore Abelian.