If $G$ is a finite group and $g \in G$, then $O(\langle g\rangle)$ is a divisor of $O(G)$

Question

Does this result mean:

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

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 sync — Soham, Sep 03 '12 at 15:40
great, so if that is so, can you give me an example of it. Preferably taking a set in complex numbers. — Soham, Sep 03 '12 at 15:46
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?. — lhf, Sep 03 '12 at 16:19
Related to http://math.stackexchange.com/questions/28317/how-to-prove-the-following-about-a-group-g — lhf, Sep 03 '12 at 16:29
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! — Derek Holt, Sep 03 '12 at 19:23

Thomas · Accepted Answer · 2012-09-03T15:53:51.437

3

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.

edited Sep 03 '12 at 15:53

answered Sep 03 '12 at 15:48

Thomas

43,555

order = 3 and yes, proving it as a subgroup is easy, I guess. – Soham Sep 03 '12 at 15:59
Additionally, if O(H) = m/gcd(m,i) here m = 15, but what is i? The generator is $z^3$ isnt it? – Soham Sep 03 '12 at 16:07
You are right: the generator for $H = \langle w \rangle$ is $z^3$. I am not sure that I understand your formula. – Thomas Sep 03 '12 at 16:10
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) – Soham Sep 03 '12 at 16:13
Ok, so you have $m =15$, $i = 3$, so the order of $\langle w\rangle = \frac{15}{\gcd(15,3)} = \frac{15}{3} = 5$. – Thomas Sep 03 '12 at 16:17
Of course, how asinine of me, for a moment I was thinking gcd(15,3) is 1 – Soham Sep 03 '12 at 16:19

score 2 · Answer 2 · edited Apr 13 '17 at 12:58

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.

If $G$ is a finite group and $g \in G$, then $O(\langle g\rangle)$ is a divisor of $O(G)$

2 Answers2

Linked