Cyclic Group of Order 6

Theorem
Let $$C_n$$ be the cyclic group of order $$n$$.

Then:
 * $$C_2 \times C_3 \cong C_6$$;
 * $$C_6$$ is the internal group direct product of $$C_2$$ and $$C_3$$.

Proof

 * $$C_2 \times C_3 \cong C_6$$:

Follows directly from Group Direct Product of Cyclic Groups, as $$2 \perp 3$$.


 * $$C_6$$ is the internal group direct product of $$C_2$$ and $$C_3$$:

Let $$\left({C_6, \circ}\right) = \left \langle {x} \right \rangle$$, let $$\left({C_2, \circ}\right) = \left \langle {x_2} \right \rangle$$, and let $$\left({C_3, \circ}\right) = \left \langle {x_3} \right \rangle$$.

From All Subgroups of Abelian Group are Normal, $$C_2 \triangleleft C_6$$ and $$C_3 \triangleleft C_6$$.

We can factorise the elements of $$C_6$$ thus:

$$e = e \circ e, x = x^3 \circ \left({x^2}\right)^2, x^2 = e \circ x^2, x^3 = x^3 \circ e, x^4 = e \circ \left({x^2}\right)^2, x^5 = x^2 \circ x^3$$

Hence the result.