# Homework Help: How many homomorphisms are there

1. May 15, 2006

### JFo

How many homomorphisms are there from $S_5$ to $\mathbb{Z}_5$?

Well there is at least one, the trivial homomorphism, ie: every element of S_5 gets mapped to 0.

I have a feeling that this is the only homomorphism but am having trouble proving that no other homomorphism could exist. Any suggestions?

I know that every nonzero element of Z_5 has order of 5, and for a non-trivial homomorphism to exist, there needs to be some element of S_5 (not equal to the identity) that gets mapped to a nonzero element of Z_5.

Last edited: May 15, 2006
2. May 15, 2006

### SpongeBobRhombusHat

I believe that there are other homomorphisms. Think about the subgroups of S_5. Does it have any cyclic subgroups of order 5?

SBRH

3. May 15, 2006

### Hurkyl

Staff Emeritus
But that only gives you a map in the other direction: from the cyclic group into the permutation group.

What are the generators of S_5?

4. May 15, 2006

### SpongeBobRhombusHat

Yeah, yeah, yeah...

I just realized my silly mistake and came back to announce its sillyness!

SBRH

5. May 15, 2006

### JFo

Im thinking the generators of S_5 are all the transpositions of S_5. Is this correct?

Im working out a proof, here's what I have so far:

proof:
Let $\phi : \mbox{S}_5 \rightarrow \mathbb{Z}_5$ be a nontrivial homomorphism. Now $|\phi(\mbox{S}_5)|$ divides $|\mathbb{Z}_5|$, and thus must be 1 or 5. It can't be 1 since phi is nontrivial and thus must be 5, hence phi is onto.
Let H = Ker($\phi$). We can form the factor group $\mbox{S}_5 / H$ which is isomorphic to $\mathbb{Z}_5$. This implies that $|H| = 24$. So for each element x of Z_5 there are 24 distinct elements of S_5 that map to x.

Ok Im stuck again

Last edited: May 15, 2006
6. May 15, 2006

### Hurkyl

Staff Emeritus
Yes. Every element of S_5 is a product of transpositions.

Are there any 24-element subgroups of S_5? Normal ones?

Last edited: May 15, 2006
7. May 15, 2006

### JFo

No. There are no 24-element subgroups of S_5 that are normal. Thanks much for your help Hurkyl!
I was wondering.. is there a way, say if your looking at a subgroup diagram of S_n for some n, to tell whether a given subgroup of S_n is normal?

Also, is there a way to figure out which subgroups are conjugate without computing every conjugation of that group?

8. May 15, 2006

### Hurkyl

Staff Emeritus
Incidentally, looking at the generators is just as easy. What is the image of any transposition?

(I would have said easier, but I suppose if you really understand this stuff, knowing that there are no 24-element subgroups of S_n is probably of comparable difficulty)

Isn't it the case that A_n is the only normal subgroup of S_n, for n > 4?

9. May 16, 2006

### JFo

I suppose if $\tau$ is a transpostion, then $\tau^2 = i$, thus $\phi(i) = \phi(\tau^2) = \phi(\tau)^2 = 0$. This implies $\phi(\tau) = 0$ since there are no nonzero elements of order 2 in Z_5. Ahhh... this then implies that that all of S_5 gets mapped to 0 since every permutation can be written as a product of transpositions and phi is a homomorphism. Is this the correct reasoning?

You're right, this is an easier solution.

Last edited: May 16, 2006
10. May 16, 2006

### Hurkyl

Staff Emeritus
Yes, that is the line of thinking I originally had. (Until I was inspired by your computation of |H|)