# Properties of finite groups

1. May 4, 2005

### 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?

Last edited: May 4, 2005
2. May 4, 2005

### Hurkyl

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

3. May 4, 2005

### matt grime

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

4. May 4, 2005

### Zaare

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.

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...

5. May 4, 2005

### Hurkyl

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

6. May 4, 2005

### Zaare

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

7. May 4, 2005

### Hurkyl

Staff Emeritus
"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."

8. May 4, 2005

### 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.

9. May 4, 2005

### 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.

10. May 4, 2005

### Hurkyl

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

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

Then G is cyclic?

11. May 4, 2005

### Zaare

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

I'm sorry about the poor specification.

Last edited: May 4, 2005
12. May 4, 2005

### 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.