Subgroups and prime order elements

[wakko101]

The question:

Let n > 1 be a fixed integer and let G be a group. If the set H = {x in G : |x| = n} together with the identity forms a subgroup of G, what can be said about n?

I know that n must be prime, but I can't figure out why that would be. The elements of h only have order 1 or n and no others and the order of an element divides the order of the (sub)group. Are we making the assumption that since there are no other divisors of the group order, n must be prime? Or is it something else...

Any insight would be appreciated.

Cheers,
W. =)

 

[Eynstone]

No - take integers modulo 12 (under addition) with n=4. {0,3,6,9} is a subgroup.

 

[rs1n]

No - take integers modulo 12 (under addition) with n=4. {0,3,6,9} is a subgroup.

Is this supposed to be a counterexample? Did you account for the fact that |6|=2 since 6+6=0?

 

[rs1n]

The question:

Let n > 1 be a fixed integer and let G be a group. If the set H = {x in G : |x| = n} together with the identity forms a subgroup of G, what can be said about n?

I know that n must be prime, but I can't figure out why that would be. The elements of h only have order 1 or n and no others and the order of an element divides the order of the (sub)group. Are we making the assumption that since there are no other divisors of the group order, n must be prime? Or is it something else...

Your intuition about n being prime is correct. As for your assumption question --- it is not an assumption; it is what you have to prove!

Try a proof by contradiction. That is, suppose n = ab with a>1 and b>1. Now consider any element h in H. Since |h|=n=ab, we have $h^{ab}=(h^a)^b$. Can you see how to get a contradiction out of this?