Thread: Abstract Algebra Help!!!

Hey im preparing for my second exam and I need major help with these problems. I would greatly appreciate any help, hints or answers, that could help me understand how to do these problems. So that I can do well thanks!!!

G denotes a group

1. Let |G|=p*g*r with different primes p, g, r, and let H,K<=G with |H|=p*g, |K|=g*r. Show:|H (intersection) K|=g.

2. Let H be the normal subgroup G, a in G and |H|=10. The element aH of the group G/H has order 3; what are the possibilities for |a|? Explain.

3. Let G be a group and G' be the subgroup generated by S={x^-1*y^-1*x*y| x,y in G}. Then:

a. Prove that G' is normal in G.
b. Prove that G/G' is abelian.
c. If G/N is abelian, prove that G'<=N.
d. Prove that if H<=G and G'<=H then H is a subgroup of G.

4. Suppose that G is an abelian group of order 120 and that G has exactly 3 elements of order 2. Determine the isomorphism class of G. Explain.

Originally Posted by GreenandGold

1. Let |G|=p*g*r with different primes p, g, r, and let H,K<=G with |H|=p*g, |K|=g*r. Show:|H (intersection) K|=g.

If are subgroups of a finite group then .

Now since is a subgroup of and it means its order divides and . Therefore, . We cannot have for that would imply . A contradiction.

Therefore, the only possible case is that .

2. Let H be the normal subgroup G, a in G and |H|=10. The element aH of the group G/H has order 3; what are the possibilities for |a|? Explain.

If in the group it means and therefore . But . Therefore, . Since it means the possible values for are .

Originally Posted by GreenandGold

3. Let G be a group and G' be the subgroup generated by S={x^-1*y^-1*x*y| x,y in G}. Then:

a. Prove that G' is normal in G.
b. Prove that G/G' is abelian.
c. If G/N is abelian, prove that G'<=N.
d. Prove that if H<=G and G'<=H then H is a subgroup of G.

For define then show that .
Therefore, we see that is a normal subgroup.

Since this group is normal we can form the factor group . Let . To show it is necessary and sufficient to prove that . This is true by definition of .

If is abelian it means and so . This is true for all . Therefore, .

Originally Posted by GreenandGold

4. Suppose that G is an abelian group of order 120 and that G has exactly 3 elements of order 2. Determine the isomorphism class of G. Explain.

Let be the Sylow -subgroup of order .

Consider, . This group has order and therefore because that is the only group up to isomorphism.

Since and it is abelian implies that must be isomorphic to one of the following groups: .
Elements of order must be contained in . Therefore, by hypothesis, we are told that has exactly elements of order .
The first two groups (on the list) do not satisfy this requirement.
However, has that requirement with elements .

Therefore, .

Now we "cross multiply" and we get, .