Groups containing no proper subgroup

[hgj]

1. Describe all groups G which contain no proper subgroup.

This is my answer so far:
Let G be a such a group with order n. Then the following describe G:
(a) Claim that every element in G must also have order n.
Proof of this: If this wasn't true, the elements of lower order (elements of order m where m<n) would generate smaller subgroups, thus creating proper subgroups. (Though I'm not sure what to say about elements of order greater than n--is that possible?)
(b) Claim G must also be prime.
Proof of this: If G is not prime, then n = k*m for some integers k,m < n. Also, if G is not prime, then I think a^k would have order m, m < n (but I'm not sure why this second part is true).




2. Let a,b be elements of an abelian roup of orders m,n respectively. What can you say about the order of their product?

My answer (in progress):
If the orders of a and b are m and n, respectively, then a^m=1 and b^n=1. Then a^m=b^n and a^mb^n=1. Let m<n. Then a^mb^n=(ab)^mb^(^n^-^m^)=1.
I'm really not sure where to go from here.

3. (related to 2.) Show by example that the product of elements of finite order in a nonabelian group need not have finite order.

My answer (in progress):
A nonabelian group can be infinite in order, meaning their are infinitely many elements in the group. But the individual elements in the group can still have finite order. Hence, even if the elements have finite order, there are infinitely many of these such elements, so their product does not have to have finite order.
I'm not convinced of this answer, so any advice would be very appreciated.

Also, any input on any of the questions would be helpful. Thanks!

 

[AKG]

Have you done Lagrange's theorem? If so, then you'll know that no element of a group of order n can have order greater than n. You are more or less right that every element will have to have order n, what you should actually say is that every non-identity element must have order n. Identity will generate its own little subgroup, but I guess you could consider it an improper subgroup. Again, if you are familiar with Lagrange's theorem, then you'll also know why n must be prime. You will be able to say that these groups are cyclic, and hence abelian.

Well you're not really on the right track for 2. Try multiplying (ab) by multiples of m.

I'm not sure how 3 is related to 2. However, the product of two reflections in lines passing through the origin is a rotation. Any reflection has order 2, but if you choose your lines right, the resulting rotation can have infinite order, just make sure that it is a rotation that is not a rational multiple of pi.

 

[hgj]

We have not done Lagrange's Theorem yet, though I've looked at it and wish we had...it seems it would be helpful.


I'm still not sure what to do for the second problem. I don't see what multiplying (ab) my multiples of m would do for me.

 

[hgj]

Okay, with the second problem about the order of the product of ab, I think it's that the order of ab divides the lcm(m,n) where m and n are the orders of a and b. I'm not sure how to prove this though.

 

[AKG]

lcm(m,n) = kmn for some k in {1, 2, 3, ...}, agreed? So:

(ab)^lcm(m,n)
= (ab)^kmn
= a^kmnb^kmn ... since the group is abelian
= (aⁿ)^km(b^m)^kn
= e^kme^kn
= e

Now there's a stronger statement you can make regarding the relationship between lcm(m,n) and the order of ab, and making this statement will solve the problem. Try going about this problem by looking at the subgroup generated by a and b, <a, b>.

 

[hgj]

if (ab)^l^c^m^(^m^,^n^) = e, doesn't that mean that the order of ab is the lcm(m,n)? Then, because a number always divides itself, that would explain why the order of ab divides the lcm(m,n). Is that right?

 

[AKG]

Well e¹⁰⁰ = e, but 100 is not the order of e. The above shows that the order of (ab) divides lcm(m,n), not that it is lcm(m,n). The fact is that it is lcm(m,n) but I'm not sure how to prove it actually. You have to prove that if you raise ab to some power less than lcm(m,n) then you cannot get e.