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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 [tex]a^m=1[/tex] and [tex]b^n=1[/tex]. Then [tex]a^m=b^n[/tex] and [tex]a^mb^n=1[/tex]. Let m<n. Then [tex]a^mb^n=(ab)^mb^(^n^-^m^)=1[/tex].

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!