How many homomorphisms are there

  • Thread starter JFo
  • Start date
  • #1
JFo
92
0
How many homomorphisms are there from [itex]S_5[/itex] to [itex]\mathbb{Z}_5[/itex]?

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:

Answers and Replies

  • #2
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
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
19
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
Yeah, yeah, yeah...

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

Ignore my previous reply.

SBRH
 
  • #5
JFo
92
0
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 [itex] \phi : \mbox{S}_5 \rightarrow \mathbb{Z}_5[/itex] be a nontrivial homomorphism. Now [itex] |\phi(\mbox{S}_5)| [/itex] divides [itex]|\mathbb{Z}_5|[/itex], 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([itex]\phi[/itex]). We can form the factor group [itex]\mbox{S}_5 / H[/itex] which is isomorphic to [itex]\mathbb{Z}_5[/itex]. This implies that [itex]|H| = 24 [/itex]. 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:
  • #6
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
19
Im thinking the generators of S_5 are all the transpositions of S_5. Is this correct?
Yes. Every element of S_5 is a product of transpositions.


This implies that |H| = 24.
Are there any 24-element subgroups of S_5? Normal ones?
 
Last edited:
  • #7
JFo
92
0
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
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
19
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
JFo
92
0
I suppose if [itex] \tau [/itex] is a transpostion, then [itex] \tau^2 = i[/itex], thus [itex]\phi(i) = \phi(\tau^2) = \phi(\tau)^2 = 0[/itex]. This implies [itex]\phi(\tau) = 0[/itex] 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.
Thanks again for your help!
 
Last edited:
  • #10
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
19
Yes, that is the line of thinking I originally had. (Until I was inspired by your computation of |H|)
 

Related Threads on How many homomorphisms are there

Replies
3
Views
1K
Replies
5
Views
1K
Replies
5
Views
6K
Replies
15
Views
3K
J
Replies
4
Views
1K
  • Last Post
Replies
3
Views
633
Replies
0
Views
1K
  • Last Post
Replies
4
Views
2K
Top