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Proving this can be done as follows: consider a finite group G and elements $g_i \in G$ for some integer $i$. Now consider $\langle g_i \rangle = \{g_i^n: n\geq 0\}$, a generator. It can be proved that $\langle g_i \rangle \leq G$ and that the order of $g_i$ is equal to the order of $\langle g_i \rangle$, so $|\langle g_i \rangle| \leq |G|$.

We can now use Lagrange's theorem which states that if $H \leq G$ then $|H|$ divides $|G|$ and we're done. But...

Is there a simpler way to prove this fact?

sequence
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  • The order of an element is the order of the cyclic group it generates, which divides the order of the group by Lagrange. I think most other proofs boil down to this. –  Sep 26 '15 at 22:42
  • That's one of my pet questions: http://math.stackexchange.com/questions/28332/is-lagranges-theorem-the-most-basic-result-in-finite-group-theory – lhf Oct 10 '15 at 01:10
  • See also http://math.stackexchange.com/a/1542608/589. – lhf Nov 26 '15 at 12:13

2 Answers2

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Yes but it is almost same with the proof of Lagrange theorem. (as a pattern)

Choose $x\in G$, think the set $\{x,gx,g^2x,...,g^{n-1}x\}$. As you see this set has $n$ elements. Choose another $y$ which is not the set and create the set $\{y,gy,g^2y,...,g^{n-1}y\}$. ... You will see that $n$ divides order of $G$.

mesel
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  • "You will see that $n$ divides order of $G$." How? This is just what we're trying to prove, I don't see how it follows from what you said. – nog642 Jan 14 '24 at 04:57
  • The above argument shows that we can "consume the set" by removing n elements repeatedly, so the cardinality of the set must be divisible by n. – mesel Jan 14 '24 at 12:45
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Fix an index $i \in \mathbb{N}$. Let $\langle g_{i}\rangle$ act on $G$ by left multiplication. Notice that the orbits under the action of $\langle g_{i}\rangle$ partition the group $G$. Let their number equal $k$. Now, notice that all orbits have cardinality $|\langle g_{i}\rangle|$ (Prove that for every $x \in G$ the map $\langle g_{i}\rangle \rightarrow \mathcal{O}(x)$ ,where $\mathcal{O}(x)$ is the orbit of $G$ that contains $x$, defined by $g_{i}^{n} \mapsto g_{i}^{n}x$ is a bijection). So, $$|G|=\sum_{j=1}^{k}|\mathcal{O}_{j}|=\sum_{j=1}^{k}|\langle g_{i}\rangle|=k|\langle g_{i}\rangle|$$ and as a result the order of $g_{i}$ divides the order of the group $G$.

(It is practically the same answer as $\textbf{mesel's}$ but more detailed. I hope that it helped you.)

Martin Sleziak
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vasmous
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