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

[e(ho0n3]

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?

 

[e(ho0n3]

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 = xⁱy^j and b = x^ky^m be elements of HK.

Then ab^-1 = xⁱy^j-mx^-k. Let n be such that k + n = i. Then xⁱy^j-mx^-k = x^k(xⁿy^j-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.

 

[Dick]

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.

 

[e(ho0n3]

Coming from you, that's not a good sign. Can you tell me the flaw in my proof?

 

[Dick]

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?

 

[e(ho0n3]

That's exactly what I was thinking. I realize now that it is wrong. Oh well.

 

[Dick]

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.