Cyclic group problem

**tttcomrader** · September 29th 2007, 08:28 PM

Let G be a group of order 25. Prove that G is cyclic or g^5=e for all g in G.

**Rebesques** · September 30th 2007, 03:35 AM

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.

**tttcomrader** · September 30th 2007, 08:14 AM

Oh, man, this problem is easy. I really should have gotten it.

Thanks.

**ThePerfectHacker** · September 30th 2007, 08:18 AM

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.

**Rebesques** · October 2nd 2007, 10:32 AM

I don't see where you disagree PH

**ThePerfectHacker** · October 2nd 2007, 05:15 PM

Originally Posted by Rebesques

I don't see where you disagree PH

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

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