Any group which is of prime order is a cyclic group

Question

I don't know how to prove this:

Any group which is of prime order is a cyclic group.

What fact should I use to prove this?

http://math.stackexchange.com/questions/106163/show-that-every-group-of-prime-order-is-cyclic — , Nov 16 '14 at 11:40

score 4 · Accepted Answer · answered Nov 16 '14 at 11:35

4

Hint: Suppose $|G|=p$ where $p$ is a prime.

Suppose, $a\in G$. What can you say about $|a|$?

Note $|.|$ denotes order.

answered Nov 16 '14 at 11:35

Swapnil Tripathi

the order of $a$ must divide the order of group using the lagrange's theorem.. hence it can be $1$ or $p$.. but why shouldn't it be $1$.. – spectraa Nov 16 '14 at 11:44
@spectraa: But can all the $p$ (>1) elements of group $G$ have order $1$? – Swapnil Tripathi Nov 16 '14 at 11:57
but we are considering just $1$ element.. – spectraa Nov 16 '14 at 11:59
Yes, we are. But I'm trying to prove that there will be an element of order $p$. There will be an element such that the order $\ne$1. Since we reach a contradiction if this is not the case. – Swapnil Tripathi Nov 16 '14 at 12:01
1

@spectraa Just change it in “Suppose $a\in G$, $a\ne1$.” – egreg Nov 16 '14 at 12:02
@Swapnil I understood ,you mean that we are proving for any general element $a$ of $G$ ,so we should not consider special case of $a=1$..is it what you mean to say.. – spectraa Nov 16 '14 at 12:04
@spectraa Yes. But do you understand the reasoning? :) – Swapnil Tripathi Nov 16 '14 at 12:07
1

@spectraa: As egreg says you can directly assume $a\ne \text{identity}$ (since identity is unique) and you're done. – Swapnil Tripathi Nov 16 '14 at 12:14

score 2 · Answer 2 · answered Nov 16 '14 at 11:35

2

Take a nonzero element a in the group.
Consider the group generated by a, which is of course cyclic.
Try to recall what you can say about the order of a subgroup relative to the order of the group itself, and what this implies in your case.

answered Nov 16 '14 at 11:35

quid

2 Answers2