Proving the Simplicity of a Group of Order 36 Using Sylow Subgroups

[Bachelier]

Well it isn't.
I'm trying to prove it.

So I assume there is more than one sylow-3 subgroup, each has order 9, we have 4 of them, now their intersection is either e or has order 3.

if it is e, then we have 32 elements in these subgroups besides e (33rd)

then assume we have more than one sylow-2 subgroup, we should have 3, but then their intersection has order 1, or 2.

So now it works for all cases except when the intersection of the sylow 3 subgroups is 3 and that of the sylow 2-subgrps is 1, then I get the magic number 36.

What should I be looking for in this proof. I was advised there is a way to prove it using this way without the appeal to homomorphisms to Sn.

 

[mathwonk]

if you check out my math notes for 843-4-5 on my website you will find proofs that all groups of non prime order n different from 60 are not simple for n < 168.843-1. #9.

http://www.math.uga.edu/%7Eroy/843-1.pdf

 

[Bachelier]

if you check out my math notes for 843-4-5 on my website you will find proofs that all groups of non prime order n different from 60 are not simple for n < 168.


843-1. #9.

http://www.math.uga.edu/%7Eroy/843-1.pdf

Thanks, I am going to check it now.
btw I corrected my answer to the intersection of sylow p-subgroups. I didn't know that there's a theory on these. The Sylow intersection.

so the ord(sylow intersection) divides the original group but must NOT be equal or larger to the order of any sylow-p-group (which is the same for all since they are all isomorphic).

 

[Bachelier]

when it comes to group theory, Hungerford's introductory textbook is truly weak. Not to confuse with his Graduate textbook Algebra though.

 

[Deveno]

suppose we have the "bad case" where we have the intersection of the sylow 3-subgroups is non-trivial. let's call one of these sylow 3-subgroups P.

this means that $\bigcap_{x \in G}xPx^{-1}$ is a subgroup of order 3. show this is a normal subgroup of G, and then G is not simple.

there is also another way:

suppose we have 2 sylow-3-subgroups H,K with non-trivial intersection. what are possible orders for N(H∩K)?

(first rule out 3 and 9. from the first sylow theorem we know there is a subgroup of G of order 9, that H∩K is normal in. this rules out 3. now rule out 9, since...?)

 

[Bachelier]

suppose we have the "bad case" where we have the intersection of the sylow 3-subgroups is non-trivial. let's call one of these sylow 3-subgroups P.

this means that $\bigcap_{x \in G}xPx^{-1}$ is a subgroup of order 3. show this is a normal subgroup of G, and then G is not simple.

there is also another way:

suppose we have 2 sylow-3-subgroups H,K with non-trivial intersection. what are possible orders for N(H∩K)?

(first rule out 3 and 9. from the first sylow theorem we know there is a subgroup of G of order 9, that H∩K is normal in. this rules out 3. now rule out 9, since...?)

I'm inclined to think that our subgroup $\bigcap_{x \in G} \ xPx^{-1}$ is nothing but the original intersection subgroup. The reason: xPx^-1 is nothing but another sylow-3-subgrp. Thus the order is 3. Also since for all x in G, if we take $x \bigcap x^{-1}$ then we end up in the same grp hence it's normal.

 

[Bachelier]

suppose we have 2 sylow-3-subgroups H,K with non-trivial intersection. what are possible orders for N(H∩K)?

(first rule out 3 and 9. from the first sylow theorem we know there is a subgroup of G of order 9, that H∩K is normal in. this rules out 3. now rule out 9, since...?)

How can we have 2 Sylow 3-subgroups and $2 \not \equiv 1 \mod{3}$

 

[Deveno]

How can we have 2 Sylow 3-subgroups and $2 \not \equiv 1 \mod{3}$

no, i don't mean we have ONLY 2. i mean suppose we have 4. pick any two of them, call them H and K. they have a non-trivial intersection. what is the order of the normalizer of this intersection?

(hint: any subgroup of G of order 9 is abelian. thus H∩K is normal in both H and K, which means that |N(H∩K)| > 9. since both H and K are in N(H∩K), this means that 9 divides N(H∩K). so we have 9 divides |N(H∩K)| and |N(H∩K)| divides 36. what possibilities does this give?)

I'm inclined to think that our subgroup $\bigcap_{x \in G} \ xPx^{-1}$ is nothing but the original intersection subgroup. The reason: xPx^-1 is nothing but another sylow-3-subgrp. Thus the order is 3. Also since for all x in G, if we take $x \bigcap x^{-1}$ then we end up in the same grp hence it's normal.

the group $\bigcap_{x \in G} \ xPx^{-1}$ is called the normal hull of P. it is always a normal subgroup of G. see this thread: https://www.physicsforums.com/showthread.php?t=555966

 

[Bachelier]

First let me ask this please, regarding the normalizer, is there a formula for its order?

 

[Deveno]

First let me ask this please, regarding the normalizer, is there a formula for its order?

in general, no. but for any subgroup H of G, we have H ≤ N_G(H) ≤ G. so normalizers can be used when we need to find "bigger" subgroups of G containing H.