Can the Direct Sum of Cyclic Groups Determine the Properties of Finite Groups?

[Zaare]

Assume G is a finite group and H = \left\{ {g \in G|g^n = e} \right\} for any n>0. e is identity.
I have been able to show that if G is cyclic, then H has at most n elements.
However, I can't go the other way. That is, assuming H has at most n elements, I haven't been able to say anything about whether G is cyclic, abelian or neither.
Any suggestions?

 

[Hurkyl]

Can you add an element to G without adding an element to H?

 

[matt grime]

Look at D_3=S_3 the nonabelian group of order 6. How many elements of order 3 does it have?

 

[Zaare]

Can you add an element to G without adding an element to H?

Since G is a finite group, then every element in G must equal identity for some n. That means that for some n the element must be added to H. So I think the answer is no. But I can't make any connection to my problem.

Look at D_3=S_3 the nonabelian group of order 6. How many elements of order 3 does it have?

If I haven't made any mistakes, it has 2 elements of order 3 and 3 elements of order 2. But that means that H has 3 elements for n=2 (which does not agree with the assumption that H has at most n elements), doesn't it?

I'm quite confused now...

 

[Hurkyl]

You seem to be treating n as both a constant and a variable -- that might be the source of confusion.

 

[Zaare]

I don't mean to treat n as a variable, only as an unknown constant. Where do I treat it as a variable?

 

[Hurkyl]

"Since G is a finite group, then every element in G must equal identity for some n. That means that for some n the element must be added to H."

 

[matt grime]

S_3 is a group that has fewer than 3 elements of order 3. How does that not prove useful unless you're being odd about n: at least quantify it: there exists an n, or "for all n".

It apparently seems you wish it to be "for all n", which you didn't bother to specify.

 

[Zaare]

But a certain value of n defines a certain H, so I have to consider the different values n can take.

What I mean is for any n.

 

[Hurkyl]

So, what you want to prove (or disprove) is that if:

For all n > 0, H_n has no more than n elements

Then G is cyclic?

 

[Zaare]

Yes, and if it's not cyclic: Is it "at least" abelian?

I'm sorry about the poor specification.

 

[mathwonk]

try the direct sum of the cyclic groups of orders 2,3,5,7,11,13,17,19,23,29. and look for elements of order 2.