# Abelian Group Element Orders

I want to show that if a finite abelian group has elements of order $m$ and $n$ then it will have an element of order $\text{lcm}(m,n)$.

First I proved the lemma if $a$ has order $m$ and $b$ has order $n$ with $m,n$ coprime, then $ab$ has order $mn$. This is because $\langle a \rangle$ and $\langle b \rangle$ are not subgroups of each other (because they are both nontrivial and the order of one does not divide the order of another, which is a consequence of something being a subgroup) which implies $a^{i} \in \langle b \rangle$ iff $i \equiv 0 \pmod n$, and similarly $b^{i} \in \langle a \rangle$ iff $i \equiv 0 \pmod m$. Using that we deduce that $(ab)^i = a^i b^i = 1$ iff $i \equiv 0 \pmod {mn}$.

So if $a$ was an element of order $m$ and $b$ and element of $n$ with $g = \gcd(m,n) \not = 1$ I thought that $\text{lcm}(m,n) = \frac{m}{g}n$ so and $\frac{m}{g},n$ are coprime so I should construct an element (from $a$) with order $\frac{m}{g}$ then conclude the theorem by the lemma. For the construction I think it's just $a^g$.

I just have a nagging doubt about the correctness of the second proof, did I miss some important detail? Also if there are any neater ways to prove this (which don't depend on the structure theorem) I would like to learn them too.

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I am no group theory specialist, but this proof looks fine. –  dtldarek Dec 11 '12 at 0:50

Everything you said seems fine. Using additive notation, you only need to consider the subgroup $sa + tb$ of $G$. This group contains the element $a+b$ which has the property
$$\hbox{lcm}(m,n) \cdot (a + b) = 0$$
so the order of $a+b$ is at most $\hbox{lcm}(m,n)$.
Moreover the order must be equal to $\hbox{lcm}(m,n)$, because any smaller order could be divisible by both $m$ and $n$.
You mean be divisible by both $m$ and $n$. And we must justify the lemma that the order of $a+b$ must be divisible by the orders of $a$ and $b$. :-) –  anon Dec 11 '12 at 1:03