$C_m \times C_n$ isomorphic to $C_{mn}$

Question

Let $m, \, n$ be coprime integers.

(a) Let $G$ be an abelian group containing elements of orders $m$ and $n$. Prove that $G$ contains an element of order $mn$.

(b) Deduce from part (a) that the group $C_m \times C_n$ is isomorphic to $C_{mn}$.

Now I've already done part (a) and shown that the element whose order is $mn$ is in fact the element $xy$ where ord$(x) = m$ and ord$(y) = n$. I've also deduced from part (a) that the group containing $x$ and $y$ where $<x>=C_m$ and $<y>=C_n$ does indeed have an element of order $mn$ and that element is $xy$ with $<xy>=C_{mn}$ however, I don't really know where to go from there. Any guidance would be appreciated.

Answer 1

Hint : Deduce from part (a) that $C_{m}\times C_n$ contains an element of order $mn$, or equivalently, a subgroup isomorphic to $C_{mn}$. Then show that this subgroup is actually the whole group (sub-hint : how many elements does each group have?)

Answer 2

Let $C_m=\langle a\rangle$ and $C_n=\langle b\rangle$. Then $\langle a\rangle \langle b\rangle$ is subgroup of $\langle a, b\rangle$ with same order. Then $C_m\times C_n\cong \langle a\rangle\langle b\rangle$ and $\langle a, b\rangle\cong C_{mn}$. So, $C_m\times C_n\cong \langle a\rangle\langle b\rangle=\langle a, b\rangle\cong C_{mn}$.

Why $\langle a\rangle\langle b\rangle=\langle a, b\rangle$ is true?

Equality has set propert as follow: If $\mid A\mid= \mid B\mid$ and $A\subseteq B$ then $A=B$.