Finding Surjective Homomorphisms from Symmetric Groups to Cyclic Groups

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Homework Help Overview

The discussion revolves around finding positive integers r such that there exists a surjective homomorphism from the symmetric group S_n to the cyclic group C_r for some n. The original poster expresses uncertainty about how to approach the problem in group theory.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the implications of the relationship between the orders of the groups involved, particularly questioning what occurs when r is greater than n. There is a discussion about the conditions under which a surjective homomorphism can exist, including references to normal subgroups and the isomorphism theorems.

Discussion Status

Some participants have provided insights regarding the structure of groups and the properties of homomorphisms, while others are still seeking clarity on how to identify suitable values of r for given n. The conversation reflects a mix of understanding and confusion, with no explicit consensus reached.

Contextual Notes

There is mention of the importance of group orders and the divisibility of elements, as well as the original poster's acknowledgment of the need to find all r for some n that would yield a homomorphism from S_n to C_r.

Indran
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Hello,
I am having difficulty with the following problem in Group theory:

How do you positive integers r such that there is a surjective homomorphism from S_n (This is the symmetric group of order n) to
C_r (This is the cyclic group order r) for some n ?
I am not sure where to even start and any pointers in right direction will
be greatly appreciated.
 
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What happens when r is greater than n?
 
Groups

AKG said:
What happens when r is greater than n?

Well, on the face of it. since r>n I would have said that therecannot be a surjective relationship, but if say, S_3 ={1 2 3} and C_4 (thus r>n) is
{e, a, a^2, a^3} the order of S_3 is 3! = 6 and the order of C_4 is 4
so there could be a surjective homomorphism. I am stuck at this point as I need to find values of r that would give this homomorphism for some n.
Can you help?
 
You're ignoring many importan aspects of the structure of groups and homomorphisms. Do you know the isomorphism theorems? Your problem asks you when is there a normal subgroup H of S_n with S_n/H isomorphic to C_r. There can't be a surjective hom from S_3 to C_4 because 6/4 is not an integer. Homs alse have certain properties to do with orders of elements as well. It's impossible to send any element of S_n to any element of order 4 in any other group at all since 4 does not divide the order of any element of S_4 either.

(I presume the question did not ask you to find *all* n and r with this property - it is possible, but would require you to know too much at this stage).
 
Last edited:
Groups

matt grime said:
You're ignoring many importan aspects of the structure of groups and homomorphisms. Do you know the isomorphism theorems? You're problem asks you when is there a normal subgroup H of S_n with S_n/H isomorphic to C_r. There can't be a surjective hom from S_3 to C_4 because 6/4 is not an integer. Homs alse have certain properties to do with orders of elements as well. It's impossible to send any element of S_n to any element of order 4 in any other group at all since 4 does not divide the order of any element of S_4 either.

(I presume the question did not ask you to find *all* n and r with this property - it is possible, but would require you to know too much at this stage).

Thanks for the clear reply.
Yes, I looked up the Isomorphism theorem and now can understand what you are saying. The question does ask for *all* r for some n that will give a homomorphism from S_n to C_r. How can this be done?
 
Indran said:
{e, a, a^2, a^3} the order of S_3 is 3! = 6 and the order of C_4 is 4 so there could be a surjective homomorphism. I am stuck at this point as I need to find values of r that would give this homomorphism for some n.
Think harder.
 

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