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Does this result mean:

  1. Given any finite group, if we are able to find a cyclic group out of it (subgroup), then the order of the cyclic group will be a divisor of the original group.

If I am right in interpreting it, can one suggest an example of highlighting this? And also make me understand the possible practical uses of this result. It surely looks interesting

Thanks

Soham

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    This is a special case of Lagrange's theorem.2012-09-03
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    yes, I am sorry, I corrected it.Thanks2012-09-03
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    Yes, I understand it follows from Lagrange's theorem, the proof doesnt look complicated if one knows LT. I was just thinking if my understanding is in sync2012-09-03
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    yes, you understand correctly2012-09-03
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    great, so if that is so, can you give me an example of it. Preferably taking a set in complex numbers.2012-09-03
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    Finding a proof that does not need the full Lagrange theorem is my favorite question: [Is Lagrange's theorem the most basic result in finite group theory?](http://math.stackexchange.com/questions/28332/is-lagranges-theorem-the-most-basic-result-in-finite-group-theory).2012-09-03
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    Related to http://math.stackexchange.com/questions/28317/how-to-prove-the-following-about-a-group-g2012-09-03
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    By the way, $O(G)$ is not a standard notation for the order of $G$. It is better to use $|G|$, because $O(G)$ is often used to mean something more complicated!2012-09-03

2 Answers 2

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You have a finite group $G$ and you take any element $g\in G$. Then $\langle g \rangle$ is a subgroup of $G$. Then, as mentioned in the comment by anon, you can apply Lagrange's theorem to get the conclusion that you want.

As an example of this, you could consider the symmetric group $S_5$. You pick a random element $\sigma \in S_5$, for example $\sigma = (1, 2, 4)$. Then you get the subgroup $$ \langle \sigma\rangle = \{(1,2,4), (1, 4, 2), (1) \}. $$ Hence the order of $\langle \sigma\rangle$ is $3$, and indeed 3 is a divisor in $O(S_5) = 5! = 120$.

You ask in the comment above about an example with a subgroup of the complex numbers. Consider $z = e^{\frac{2\pi i}{15}}$. Then you have the group $G = \langle z\rangle$ (under multiplication). This group has order $15$. Can you find/write down the elements?)Now take $w= e^{\frac{2\pi i}{5}}$. Then $\langle w\rangle$ is a subgroup of $G$ of order ... ( I will let you think about that).

As an application of this someone else might have something helpful to say.

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    order = 3 and yes, proving it as a subgroup is easy, I guess.2012-09-03
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    Additionally, if O(H) = m/gcd(m,i) here m = 15, but what is i? The generator is $z^3$ isnt it?2012-09-03
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    You are right: the generator for $H = \langle w \rangle$ is $z^3$. I am not sure that I understand your formula.2012-09-03
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    Its a theorem, it says, if G is a cyclic group with order m and H is subgroup (obviously subgroup of a cyclic group is cyclic and let its generator be z^i) then O(H)= m/gcd(m,i)2012-09-03
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    Ok, so you have $m =15$, $i = 3$, so the order of $\langle w\rangle = \frac{15}{\gcd(15,3)} = \frac{15}{3} = 5$.2012-09-03
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    Of course, how asinine of me, for a moment I was thinking gcd(15,3) is 12012-09-03
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An application of this result is the formula $$ \sum_{d\mid n} \phi(d) = n $$ which can be estabilished by considering the cyclic group of order $n$: every element in this group has an order which is a divisor of $n$ and for every divisor $d$ of $n$ there are exactly $\phi(d)$ elements of order $d$.

A consequence of this formula is that finite multiplicative subgroups of a field are cyclic. In particular, the multiplicative group of a finite field is cyclic.

A simpler but very important consequence of the theorem is that groups of prime order are cyclic.