1. ## |<a>| = |a|

how to prove this:

Let G be a group and a e G an element of finite order. Then |<a>| = |a|.

Thank you!! ;]

2. Prove that $\displaystyle <a>= \{a, a^2, a^3, \cdot\cdot\cdot, a^{|a|- 1}, e\}$ where "e" is the group identity.

3. EDIT: apparently, it took me 5 minutes to reply to this. HallsofIvy's way is better than mine, methinks...

4. I am not so sure of the proof i have. Can please show me? I am not that good in proving.

5. Well, what have you written so far?

6. A proof on the corollary of the Lagrange's Theorem. It's |a| divides |G|. I just stated there that a part of the proof for the corollary is to note/recall that |<a>| = |a|. And i think i also have to show the proof for |<a>| = |a|, because maybe my classmates might ask. ;]

7. Yes, but what have you written with respect to this problem? Have you looked at what HallsofIvy suggested?

8. Im really not good in proving.

9. Yes, but you have to at least try! We're not here to do your work for you, but to help you with it. This involves, well, you doing some work...