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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- September 29th 2007, 09:28 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.

- September 30th 2007, 04:35 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. - September 30th 2007, 09:14 AMtttcomrader
Oh, man, this problem is easy. I really should have gotten it.

Thanks. - September 30th 2007, 09:18 AMThePerfectHacker
The group is not necessarity cyclic, how about .

If is cyclic then it is clearly true because and .

If is not cyclic then choose so that . Then is a subgroup of . 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 AMRebesques
I don't see where you disagree PH :o

- October 2nd 2007, 06:15 PMThePerfectHacker