Is HK a Subgroup of S_5? - Homework Statement & Equations

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Homework Help Overview

The problem involves determining whether the product of two subgroups, H and K, of the symmetric group S_5 is itself a subgroup. H is generated by the 3-cycle (1 2 3) and K by the 5-cycle (1 2 3 4 5). The question specifically asks if HK is a subgroup of S_5, with the condition that HK is a subgroup if and only if HK = KH.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss whether it is possible to show HK is a subgroup without explicitly computing HK and KH. One participant suggests using the definition of a subgroup, while another questions the validity of this approach based on the cycle structures involved.

Discussion Status

The discussion is ongoing, with participants exploring different perspectives on the proof's validity. Some express skepticism about the initial proof provided, while others acknowledge the potential for alternative methods to demonstrate that HK is not equal to KH without exhaustive computation.

Contextual Notes

There is a mention of the cycle structure of elements in HK and KH, with specific examples provided to illustrate concerns about the assumptions made in the proof. Participants are also considering the implications of these structures on the subgroup property.

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Homework Statement


Let H, K be subgroups of S_5, where H is generated by (1 2 3) and K is generated by (1 2 3 4 5). Is HK a subgroup of S_5?


Homework Equations


HK is a subgroup iff HK = KH.

The Attempt at a Solution


Is there an easy way of answering this question without computing HK and KH?
 
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Maybe the simplest thing to try is to show that HK is a subgroup directly from the definition of subgroup.

HK is obviously non-empty. Let x = (1 2 3), let y = (1 2 3 4 5) and let a = xiyj and b = xkym be elements of HK.

Then ab-1 = xiyj-mx-k. Let n be such that k + n = i. Then xiyj-mx-k = xk(xnyj-m)x-k = cd where c is in H and d is in K, since cojugating preserves cycle structure. Right? Thus, ab-1 is in HK and HK is a subgroup.
 
I don't believe that proof for a minute. The product of the two groups contains all kinds of cycle structure. Like HK contains (3,4,5). That's a 3-cycle and it's not in H. And I'll tell you another thing. (3,4,5) is not in KH. I haven't figured out any special tricks to know that yet. I did it the hard way.
 
Coming from you, that's not a good sign. Can you tell me the flaw in my proof?
 
e(ho0n3 said:
Coming from you, that's not a good sign. Can you tell me the flaw in my proof?

I can't really tell you because I don't see how it proves anything. You seem to be thinking everything in HK has the cycle structure of a disjoint three cycle and a five cycle. And the three cycle must belong to H and the five cycle must belong to K. Is that what you are thinking? I can only guess. It's not true. (1,2,3)*(1,3,5,2,4)=(2,4)*(3,5). So?
 
That's exactly what I was thinking. I realize now that it is wrong. Oh well.
 
e(ho0n3 said:
That's exactly what I was thinking. I realize now that it is wrong. Oh well.

That doesn't mean there isn't some trick you can use to show HK is not equal to KH without evaluating all 15 products in both sets (which really isn't that bad, only 8 of them are nontrivial in each set). It just means I haven't thought of one.
 

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