linzedun

We haven't done conjugate classes yet, so I don't think I can use that information to prove it.

This is what I got so far...

In the first part, I found two homomorphisms. All the elements of S4 mapped to the identity. And another one being, all the even permutations to 0 and odd permutations to 1. THe kernels being S4 and A4.

The second part I found one permutation, all elements in S4 going to the identity.

How do I prove that these are the only ones? My prof hinted that know the composition of Z2, that this are two subgroups would be a good start, but that doesn't make a ton of sense to me.

Any suggestions? Please help!!!

Find all the homomorphisms from S4 to Z2?

Find all the homomorphisms from S4 to Z3?

Where do I start on this one? I know that one is that everything can be mapped to the identity in Z2 or Z3, but there where do I go?

Help!

How do we know that the center can never be trivial? The center can be trivial when the order of the group is qp, where q and p are primes. I am confused here, would p have to be in the center if the order of G were p^2?

So I can prove that A4 has no subgroup of order 6 and then I have it proved right?

Question: Prove that every group of order p^2 (where p is a prime) is abelian.

What I started to say was this:

|G| = p^2, then |G/Z(G)| = |G|/|Z(G)| which can equal p^2, p or 1, since those are the divisors of p. If |G/Z(G)| = p, then G/Z(G) is cycic because one element has order 1 and the rest of order p, which means there is a generator, it is cyclic and then G is abelian.

If |G/Z(G)| = 1, then G=Z(G) so it much be abelian. However, the case where |G/Z(G)|=p^2 is where I get a little hung up. I know it never happens, but how can i prove it?

Question: Show that S4 has a unique subgroup of order 12.

So far I know that A4 is a subgroup of order 12. So I was trying to prove by contradiction that there was another subgroup of order 12, say G. So the (|G||A4|/the intersection of G and A4) = 24. Then I need to show something to the effect that the intersection of G and A4 has to be a subgroup of order 6 in A4 and that that is a contradiction.

This is as far as I got and my teacher says I am on the right track, but I have no ideas of where to go next, help please!

3, 6, 15, 30?

so the possible orders would be 1, 2, 5, 10?

Let H be a normal subgroup of G and let a belong to G. If the element aH has order 3 in the group G/H and H has order 10, what are the possibilities for the order of a?

Let G =Z4 external direct product Z4, H= {(0,0),(2,0),(0,2),(2,2)} and K=<(1,2)>. Is G/H isomorphic to Z4 or Z2 external direct product Z2? Is G/K isomorphic to either?

Sorry, guys, this isn't abstract but the calc board didn't seem very active... I could really use the help on this one...

Prove:

f(x): [a, b] ----> R has f'(x) continuous @ each x in [a,b] then the definite integral from a to b exists

Could someone please make a proof for this one? I am lost as to what to do.

Yes, I realize the orbits are of power P, and I know that if the orbit it 1, it is a fixed point. HOwever, I really don't know how to go about proving it any further than that?

Question: Let G be a permutation group on the finite group omega. Suppose IGI = power of p, where p is a prime. Prove that IFix(G)I = IomegaI under mod p. (I am using IGI to denote the order of G)

I know this has something to do with orbits and stabilizers. I also know how to get as far as the all Fix(G) are inside the center. But that is where I can't figure out how to go any further.

The question goes like this: Determine all cyclic groups that have exactly two generators?

I know that the answer is, Z, Z3, Z4, and Z6... however, I don't understand where that comes from, how do you know those are the only ones? Why isn't U(10) one? My test is tomorrow, I would really appreciate your help!!!