How can I prove that if G is an Abelian group with elements a and b with orders m and n, respectively, then G contains an element whose order is the least common multiple of m and n?

It’s an exercise from Hungerford’s book, but it’s not homework. I could not solve it when I take the course on groups and I think it should be easy. There is a hint that says to first prove when m and n are coprimes. I did this part. But I have no idea how to solve the general case.

Thanks.

**Answer**

Let n=∏ipnii and m=∏ipmii, where the pis are the same. Thus lcm(n,m)=∏ipmax(ni,mi)i. Let N={i:ni≥mi} and M={i:ni<mi}. Define n′=∏i∈Npnii and m′=∏i∈Mpmii. Now a′=an/n′ is of order n′, and b′=bm/m′ is of order m′. Since n′,m′ are coprime, we know that there is an element of order n′m′=lcm(n,m).

You can also prove it using the structure theorem for abelian groups, but that's an overkill.

**Attribution***Source : Link , Question Author : Álvaro Garcia , Answer Author : Yuval Filmus*