## sylow subgroups

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aguacate
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### sylow subgroups

Just had my algebra midterm, things went well, except I couldn't figure this one out:

Find a 2-Sylow subgroup of the symmetric group on 6 letters, S6.

I had no idea how to approach this, so with nothing to lose that I wouldn't lose anyway by leaving the answer blank, I started with (14), (25), (36), and all the elements they generated, then I threw in various combinations of (1425), (1436) and such, but I only got to 14 elements including the identity, and I didn't check to see if the ones I had included so far didn't already violate the subgroup axioms (or generate more than 16 elements, damn clock).

How do you go about doing this?

Note: |S6|=6!=720=16*45, and I am assuming the notation we use is standard for the permutations, but I will explain if not.

mike-l
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### Re: sylow subgroups

No subgroup of S6 has 14 elements, (by Lagrange e.g.), so you had at least 16 (can't be 15 because you had elements of order 2).

To show that it's not more than 16, I don't have any idea off hand, other than computing all combinations.. obviously not viable.

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### Re: sylow subgroups

Is this a case, can we deduce that the 2-sylow subgroup is in fact a normal subgroup?

I know how to do this if there are only 2 prime divisors of the order of the group.

If pq=n where |G|=n
Then if P is not congruent to 1 modulo Q, then P-sylow subgroup is Normal.

If we can deduce this, it might help.
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mathangelist
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### Re: sylow subgroups

I would suggest looking for subgroups of order two first, then see if you can construct a group out of those.

btilly
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### Re: sylow subgroups

aguacate wrote:Just had my algebra midterm, things went well, except I couldn't figure this one out:

Find a 2-Sylow subgroup of the symmetric group on 6 letters, S6.

I had no idea how to approach this, so with nothing to lose that I wouldn't lose anyway by leaving the answer blank, I started with (14), (25), (36), and all the elements they generated, then I threw in various combinations of (1425), (1436) and such, but I only got to 14 elements including the identity, and I didn't check to see if the ones I had included so far didn't already violate the subgroup axioms (or generate more than 16 elements, damn clock).

How do you go about doing this?

Note: |S6|=6!=720=16*45, and I am assuming the notation we use is standard for the permutations, but I will explain if not.

Well it won't be maximal unless you have something in it that is out of A6, so let's use (12) as a generator. We want another power of 4, so throw in (3456). Between those 2 you have subgroup of size 8 (have (12) or not, and from the other have the identity, (3456), (35)(46), (3654)) so we need another 2-cycle to get to 16. We want it to not accidentally generate a group that is too big, so let's pick one that looks like it interacts with what we have fairly well. How about (35)?

Obviously interactions with (12) are not an issue, we just need to verify that the subgroup generated by (3456) and (35) is of order 8. But if you can visualize it, they are both rigid permutations of a square whose corners are labelled 3, 4, 5, and 6. That group has order 8, and we have a subgroup of size at least 5, so we have the whole subgroup.

If you don't see that trick, it is not too hard to manually demonstrate that the subgroup generated by (3456) and (35) is 1, (3456), (35)(46), (3654), (35), (46), (34)(56), and (45)(36). And our full group is all of those with or without an optional (12) thrown in. So it has 16 elements.
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btilly
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### Re: sylow subgroups

3.14159265... wrote:Is this a case, can we deduce that the 2-sylow subgroup is in fact a normal subgroup?

No. In fact the 2-sylow subgroup of Sn cannot be normal for any n > 4 because if it were then its image in An would have to be normal, and An is simple.
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### Re: sylow subgroups

^Right.

Also, nice solution.
"The best times in life are the ones when you can genuinely add a "Bwa" to your "ha""- Chris Hastings

aguacate
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### Re: sylow subgroups

btilly - nice. I guess I don't see how (3456) generates the other cycles though. I haven't slept in a while.

Guess: is it because of the inverses? I don't have a solid understanding of what inverses look like yet.

skeptical scientist
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### Re: sylow subgroups

The subgroup generated by (3456) is {(3456), (3456)2, ..., (3456)n} where n is the order of the element (3456). Of course, here that's 4, so it's just {(3456), (3456)2, (3456)3, (3456)4}. Now (3456)2 = (35)(46), (3456)3=(3654), and of course (3456)4=() (the identity permutation; not sure what notation you used) since 4 is the order of the element (3456), and any element raised to its own order is the identity.

In general, the subgroup of a group G generated by a subset S of G is all elements of G which can be written as a product of elements of S and their inverses. In a finite group, we get the inverses for free since x-1=x|G|-1 is a product of |G|-1 copies of x, since x|G|=e for any x in G.
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btilly
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### Re: sylow subgroups

aguacate wrote:btilly - nice. I guess I don't see how (3456) generates the other cycles though. I haven't slept in a while.

Guess: is it because of the inverses? I don't have a solid understanding of what inverses look like yet.

Let me be painfully explicit.

(3456) does what? Maps 3 to 4, 4 to 5, 5 to 6 and 6 to 3.

(3456)(3456) does what? It maps 3 to 4 then 5. And 5 to 6 then 3. It also maps 4 to 5 then 6, and 6 to 3 then 4. So (3456)(3456) = (35)(46).

Multiply (35)(46)(3456) again and you'll find that 3 goes to 4 then 6, 6 goes to 3 then 5, 5 goes to 6 then 4 and 4 goes to 5 then 3. (I'm obviously putting the permutation to the left of the set being permuted, and applying the permutations right to left. Some people do it the other way.) So (35)(46)(3456) = (3654).

And (3654)(3456) gives you the identity permutation. As you'd hope because a 4 cycle, rotated 4 times should be the identity.

So that's how (3456) gave rise to those 4 permutations.

Similar calculations show you that (35)(3456) = (34)(56), (35)(35)(46) = (46), (35)(3654) = (36)(45), (35)I = (35). And, of course, applying (35) to those takes you back to the original.

Now we've showed that (35) applied to all 8 elements of our potential subgroup keep you in the subgroup. If we show that (3456) does as well then our subgroup is closed. (Because any member of the set is a product of things that take our subgroup back into our subgroup.) This calculation is likewise straightforward.

Now we add in (12). Since it commutes with all of the permutations that we've defined, it is easy to show that (3456), (35), and (12) form a subgroup of size 16. Which is the right size to be a Sylow-2 subgroup, and therefore is one.
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### Re: sylow subgroups

On a side note:

|G|=(p^a)(q^b)

For a=b=1 we can determine whether the sylow subgroup is Normal. Is there any extension of this for other a,b?

I have been trying to figure this out with no luck.
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tetromino
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### Re: sylow subgroups

3.14159265... wrote:On a side note:

|G|=(p^a)(q^b)

For a=b=1 we can determine whether the sylow subgroup is Normal. Is there any extension of this for other a,b?

only knowing the order of G doesn't give you enough information. consider S4 and D8 × Z/3Z, both nonabelian groups of order 24 = 233 (where by D8 I mean the symmetries of a square)
S4 contains 3 Sylow 2-subgroups, but D8 × Z/3Z has a unique (hence normal) Sylow 2-subgroup.

3.14159265...
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### Re: sylow subgroups

It must give us some information about the p-sylow subgroups.
"The best times in life are the ones when you can genuinely add a "Bwa" to your "ha""- Chris Hastings

btilly
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### Re: sylow subgroups

3.14159265... wrote:It must give us some information about the p-sylow subgroups.

It gives us lots of information about the Sylow p-subgroups. Stare at Sylow's third theorem. If the order of G is paqb then the number np of Sylow p-subgroups has to divide qb and be congruent to 1 mod p. That means that lots of times it has to be 1, and when it is 1 the Sylow p-subgroup is normal.

So very often we can tell that the Sylow p-subgroup is normal, but not always. (A counterexample has already been given.)
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### Re: sylow subgroups

Yes I know.

As I mentioned above, given that a=b=1 we can specify exactly when there will be only one p-sylow subgroup using conditions on p mod q.

My question is regarding different a or b. What conditions will force p-sylow to be normal.
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aguacate
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### Re: sylow subgroups

if qx is not congruent to 1 mod p, for all 0<x<b

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