I have to find the possible values for $o(a)$.here is my question:
Suppose $(G,*)$ is a group with $o(G)=pq$ where $p,q$ are distinct primes. For any $a\in G$ find alll possible values of $o(a)$
I can't figure it out. any one can give me a hint
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By Lagrange, it is either $1, p, q$ or $pq$, can you find an example of each? – Pax Nov 23 '15 at 11:06
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I haven't learnt the theorem. – Prabath Hewasundara Nov 23 '15 at 11:07
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Do you know that the order of a subgroup divides the order of the group? – Pax Nov 23 '15 at 11:07
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No. I don't know – Prabath Hewasundara Nov 23 '15 at 11:09
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In what context does this question arise that Lagrange's theorem is not available? – lhf Nov 23 '15 at 11:26
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I was doing some past papers of my algebra course. – Prabath Hewasundara Nov 23 '15 at 11:29
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Hint: Note that $o(a)=|\langle a \rangle|$ and use Lagrange's theorem.
Actually, all you need is this result:
If $G$ is a finite group and $a \in G$, then the order of $a$ divides the order of $G$.
The simplest proof I know is this: take $a \in G$ and consider the map $f\colon G \to G$ given by $f(x)=ax$. Consider now the orbits of $f$, that is, the sets $\mathcal{O}(x)=\{ x, f(x), f(f(x)), \dots \} = \{ x, ax, a^2x, \dots \}$. All orbits have the same number of elements because $z\mapsto zx^{-1}y$ is a bijection $\mathcal{O}(x) \to \mathcal{O}(y)$. By definition, $|\mathcal{O}(e)| = o(a)$. Hence $o(a)$ divides $|G|$ because $G$ is the disjoint union of orbits.
Going back to your problem, we conclude that the possible orders of an element in $G$ are $1,p,q,pq$.
You'll learn later that every group of order $pq$ has an element of order $p$ and an element of order $q$ (a result called Cauchy's theorem), but not necessarily an element of order $pq$ if $G$ is not abelian.
lhf
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