# Cyclic group problem

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• Sep 29th 2007, 08:28 PM
tttcomrader
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.
• Sep 30th 2007, 03:35 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.
• Sep 30th 2007, 08:14 AM
tttcomrader
Oh, man, this problem is easy. I really should have gotten it.

Thanks.
• Sep 30th 2007, 08:18 AM
ThePerfectHacker
The group is not necessarity cyclic, how about $\displaystyle \mathbb{Z}_5\times \mathbb{Z}_5$.

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

If $\displaystyle G$ is not cyclic then choose $\displaystyle x\in G$ so that $\displaystyle x\not = e$. Then $\displaystyle \left< x\right>$ is a subgroup of $\displaystyle G$. 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.
• Oct 2nd 2007, 10:32 AM
Rebesques
I don't see where you disagree PH :o
• Oct 2nd 2007, 05:15 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.