Finding Surjective Homomorphisms from Symmetric Groups to Cyclic Groups

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To find surjective homomorphisms from symmetric groups S_n to cyclic groups C_r, one must consider the structure of groups and the isomorphism theorems. A necessary condition is that the order of the symmetric group S_n must be divisible by r, as the image of a homomorphism must have an order that divides the order of the domain. For example, S_3 has an order of 6, which cannot map surjectively to C_4 since 6/4 is not an integer. Additionally, the orders of elements in S_n must also be compatible with those in C_r, meaning that elements of order r must exist in S_n. Understanding these properties is crucial for determining valid pairs of n and r for such homomorphisms.
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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