# 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

From Group Direct Product of Cyclic Groups noting that $2 \perp 3$:

$C_2 \times C_3 \cong C_6$

Let $\left({C_6, \circ}\right) = \left \langle {x} \right \rangle$, let $\left({C_2, \circ}\right) = \left \langle {x^3} \right \rangle$, and let $\left({C_3, \circ}\right) = \left \langle {x^2} \right \rangle$.

From Subgroup of Abelian Group is Normal, $C_2 \lhd C_6$ and $C_3 \lhd C_6$.

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.

$\blacksquare$