# Cyclic group problem

• September 29th 2007, 09:28 PM
Cyclic group problem
Let G be a group of order 25. Prove that G is cyclic or g^5=e for all g in G.
• September 30th 2007, 04:35 AM
Rebesques
Well, it is either cyclic or not.

Suppose it is not cyclic and choose an element g in G. The order of g, <g>, must divide the order of G, so <g> divides 25 or <g> is 1 or 5 or 25. In every case, the result holds true.
• September 30th 2007, 09:14 AM
Oh, man, this problem is easy. I really should have gotten it.

Thanks.
• September 30th 2007, 09:18 AM
ThePerfectHacker
The group is not necessarity cyclic, how about $\mathbb{Z}_5\times \mathbb{Z}_5$.

If $G$ is cyclic then it is clearly true because $G\simeq \mathbb{Z}_{25}$ and $\left< 5 \right> = \{ 0,5,10,15,20\}$.

If $G$ is not cyclic then choose $x\in G$ so that $x\not = e$. Then $\left< x\right>$ is a subgroup of $G$. It cannot (subgroup) have order 1 because it is not identity and it cannot be 25 (the full group) because we assumed it is not cyclic. Thus it must be 5 by Lagrange's theorem.
• October 2nd 2007, 11:32 AM
Rebesques
I don't see where you disagree PH :o
• October 2nd 2007, 06:15 PM
ThePerfectHacker
Quote:

Originally Posted by Rebesques
I don't see where you disagree PH :o

I do not. I wanted to be more clear to tttcomrader.