Order of elements is lcm-closed in abelian groups

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.



Let n=ipnii and m=ipmii, where the pis are the same. Thus lcm(n,m)=ipmax(ni,mi)i. Let N={i:nimi} and M={i:ni<mi}. Define n=iNpnii and m=iMpmii. 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 nm=lcm(n,m).

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

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

Leave a Comment