Proof Help: Order of element divides order of group

Question

I'm stuck in proving that the order of an element divides the order of the group.

I've already proved Lagrange's theorem that for $H<G, |H|||G|$

And I know that for any $a\epsilon G, |a|=|<a>|$

The problem is that though I know this for a fact, I need to prove that the order of the element equals the order of the cyclic subgroup it generates but how do i prove this result to finally conclude that the order of a divides G using lagrange?

Please help.

See also https://math.stackexchange.com/questions/28332/is-lagranges-theorem-the-most-basic-result-in-finite-group-theory — lhf, Dec 10 '17 at 10:31

score 1 · Answer 1 · answered Dec 10 '17 at 10:07

1

Let $n=\operatorname{ord}a$. Then $a^n=e$ and $a^k\neq e$ if $k\in\{1,2,\ldots,n-1\}$. But then $\langle a\rangle=\{e,a,a^2,\ldots,a^{n-1}\}$ and therefore $\operatorname{ord}\langle a\rangle=n=\operatorname{ord}a$.

answered Dec 10 '17 at 10:07

José Carlos Santos

427,504

Then, do we assume by definition that $= {e,a,a^2,...,a^{n-1}}$? And also, you assumed that $a^k$ does not equal e for all $k\epsilon {1,2,....3,n-1}$ because by definition n is the smallest positive integer such that $a^n = e$? Thanks – Alea Dec 10 '17 at 10:12
No, I did not assume that $\langle a\rangle={e,a,\ldots,a^{n-1}}$. But it is obvous that $\langle a\rangle\supset{e,a,\ldots,a^{n-1}}$ and since ${e,a,\ldots,a^{n-1}}$ is a group and $\langle a\rangle$ is the smallest subgroup containing $a$… The answer to your second question is affirmative. – José Carlos Santos Dec 10 '17 at 10:16
Sorry, but I don't quite understand. Does this mean I'd have to prove that $\langle a\rangle={e,a,\ldots,a^{n-1}}$ and if so, how? I'd also have to prove that $ $ is the smallest subgroup containing a but that is obvious because the group has to satisfy the closure property – Alea Dec 10 '17 at 10:21
@A.Asad By definition, $\langle a\rangle$ is the smallest subgrout containing $a$. It is easy to see that ${e,a,\ldots,a^{n-1}}$ is the smallest subgroup containing $a$ and therefore that it is equal to $\langle a\rangle$. – José Carlos Santos Dec 10 '17 at 10:40
In laymen's terms you are seeing that $a$ generates a cyclic subgroup of $G$. The order of a cyclic subgroup is the order of it's generator, and since $ < G$ you have that the order of $$ divides the order of $G$ by Lagrange. Also the order of $$ is the order of $a$.
edit: oh woops I see now your actual question seems to be how to prove $|a| = ||$.
– Boots Dec 10 '17 at 11:06

score 1 · Answer 2 · answered Dec 10 '17 at 12:07

Well, the first part of your question comes really quickly,

$$|a| = |<a>|\,\bigg|\,|G|$$

$<a>$ is a subgroup so it's order must divide the order of the group (by Lagrange)

It seems like you are really asking why $|a| = |<a>|$

To see this recall that $$<a> = \{a^k \ | \ k \in \mathbb{Z}\}$$

When $ |a| < \infty,$ $$\ a^{n_1} = a^{n_2} \iff n_1 = n_2 \mod |a| $$

$ \\ $

In particular, $\{a^k \ | \ k \in \mathbb{Z}\} = \{e=a^0,a,a^2,a^3,\ldots,a^{|a|-1}\}$

which has $|a|$ elements.

Proof Help: Order of element divides order of group

2 Answers2