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

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- Sep 29th 2007, 08:28 PMtttcomraderCyclic 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 AMRebesques
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 AMtttcomrader
Oh, man, this problem is easy. I really should have gotten it.

Thanks. - Sep 30th 2007, 08:18 AMThePerfectHacker
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 AMRebesques
I don't see where you disagree PH :o

- Oct 2nd 2007, 05:15 PMThePerfectHacker