Order of elements is lcm-closed in abelian groups

Question 1

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.

Question 2

Let $n = \prod_i p_i^{n_i}$ and $m = \prod_i p_i^{m_i}$ , where the $p_i$ s are the same. Thus $\mathrm{lcm}(n,m) = \prod_i p_i^{\max(n_i,m_i)}$ . Let $N = \{ i : n_i \geq m_i \}$ and $M = \{ i : n_i < m_i \}$ . Define $n' = \prod_{i \in N} p_i^{n_i}$ and $m' = \prod_{i \in M} p_i^{m_i}$ . Now $a' = a^{n/n'}$ is of order $n'$ , and $b' = b^{m/m'}$ is of order $m'$ . Since $n',m'$ are coprime, we know that there is an element of order $n'm' = \mathrm{lcm}(n,m)$ .

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

Order of elements is lcm-closed in abelian groups

Answer

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