Counting Orbits Sylows Theorems etc... (II)

Counting Orbits Sylows Theorems etc... (II)
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In the final two Sylow theorems that we prove, we’re going to define the Sylow subgroup to have order pk, even if that is the order of G. That is, if we let |G|=pkm, where gcd(p,m)=1, then, even if m=1, we will say that G itself is its Sylow subgroup. We aren’t losing much generality, since we already showed that every p-group of order pk, where k1 has a subgroup of order pk1. This allows us to avoid having to deal with as many cases in the proofs.

Notice that if k12, then the subgroup of order pk1 itself has a subgroup of order pk2, and on and on, until we are left with {e}, the trivial subgroup which contains only the identity element of G.

Before we can being work on the second and third Sylow theorem, we need to establish some more basic results about homomorphisms. We will also remind ourselves of some of the results we’ve already proven in earlier entries.

In what follows, let G be a group, HG a subgroup of G, and NG a normal subgroup of G. Let α:GG/N be the homomorphism taking gG to the coset gN represented by it (i.e. the natural homomorphism).

If SG is a subset of G (usually a subgroup, but the definition works for any subset), then α(S):={sN|sS}, is a subset of G/N. Similarly, if S˙G/N, then α1(S˙):={sG|α(s)S˙}.

We proved in Counting Orbits, Sylow’s Theorems, etc… that if S˙G/N is a subgroup of the factor group then α1(S˙) is a subgroup of G. We also showed that it has |S˙||N| elements (it is the disjoint union of all of the cosets represented by the elements of S˙).

We also proved in Applying Permutations to Group Theory that the image of a subgroup is a subgroup. I.e. α(H)G/N.

Now we’re going to continue our investigation of images of subgroups.

Lemma ( alpha inverse of alpha):

Let HG be a subgroup of G, and NG a normal subgroup of G. Let α:GG/N be the natural homomorphism taking gG to gNG/N. Then α1α(H)H, is a subgroup of G.

Proof

By definition it contains H. It is a group by what we’ve shown above.

QED

Lemma (alpha of alpha inverse)

Let G, and N be as above. Let KG/N be a subgroup of the factor group. Then αα1(K)=K.

Proof

Let kαα1(K), then k=kN is a coset of N, belonging to K. Then the preimage of k is the subset kN of elements of G. Then applying α to every member of that set gives us the same coset in G/N, namely, kN (which is an element of G/N and not a subset of it). This coset is in K. Therefore αα1(K)K.

Now let kK, then, as we said before k=kN, for some coset of N. Applying αα1 to that coset brings us right back to that exact same coset. Therefore Kαα1(K).

QED

What we’ve seen from the previous two lemmas is something which shows up again and again all over math. The preimage of a mapping often behaves more nicely than the image when dealing with sets.

We define α1α(H) :=HN. That’s the more usual way of defining it, but then the theorems we prove about it become slightly less intuitive.

Definition (normalizer)

Let H be a subgroup of a group G. Then

NG(H):={gG|g1Hg=H}

is called the normalizer of H in G. Notice that it is nothing other than StabG(H) under the action of conjugation. Therefore it is a subgroup of G.

Definition (H*N)

Let N be a subgroup of a group G, let HNG(N), be a subgroup of G contained within N’s normalizer. Then if we let α be the natural homomorphism mapping

NG(N)NG(N)/N

so that α(x)=xN, we get

HN:=α1α(H)

is a subgroup of NG(N).

Notice that N by definition is a normal subgroup of its own normalizer.

The definition might seem a bit odd at first, but the set is nothing more than {hn|hH,nN}. This is fairly immediate from the definition. It is also a subgroup by construction. It has just as many elements as |N| times the size of α(H). What is the size of that? Let’s do some more lemmas!

Lemma (H intersect N is normal in H):

Let H be a subgroup of G, where HNG(N). Then HNH.

Proof

Let hH and nHN, then, since HNG(N), we have that h1Nh=N, so h1nhN. Note also that nH, so then since H is a subgroup, h1nhH, therefore it is in HN.

QED

Lemma (an isomorphism theorem)

Let H be a subgroup of NG(N), then H/(HN)α(H), the subgroup α(H) of NG(N)/N. That is, they are isomorphic.

Proof

Let’s show that the mapping is well-defined. This means that it is a homomorphism, because α is a homomorphism. Let h(HN) be a coset of HN. Then α(h(HN)=hN, operating on the entire coset as a subset of H sends every element to hN. This is because (as you can check), h(HN)=hHhN, which equals HhN, i.e. the elements of the coset hN that belong to H. The mapping is thus well-defined. Next we show that it’s 1-1.

Suppose

α(h1(HN))=α(h2(HN))

Then h1N=h2N, so they belong to the same coset. Multiplying (or operating, if you want to remain more generic) the cosets on both sides of the equation on the left by h11 gives us h11h2N=N, which tells us that h11h2N. It is clearly in H, since that is a subgroup, so it must also be in the coset HN of H. Then undoing that (operating on the left by h_1), gives us

h1(HN)=h2(HN)

So our mapping is 1-1.

It is therefore an isomorphism.

Corollary (size of H*N)

Let H, G, N, etc… be as above. Then, |HN|=|H/(HN)||N| =|H||N||HN|.

Proof

It’s nothing more than multiplying through the sizes of the groups and factor groups.

Lemma (conjugation preserves subgroups)

Let H be a subgroup of a group G, then for any gG, g1Hg is a subgroup of G.

Proof

This is a straight-forward calculation. Let g1ag and g1bg be arbitrary elements of g1Hg, so then a and b are elements of H.

(g1bg)1=g1b1(g1)1

and that is equal to g1b1g, which must also be in g1Hg, since H is a subgroup. This means that (g1ag)(g1bg)1 is equal to

g1agg1b1g=g1(ab1)g

which is in g1Hg, again, because H is a subgroup. Therefore, by the subgroup test lemma, it is a subgroup.

QED

Theorem (Sylow’s second)

Let G be a group of order pkm, where p is a prime, k,m1 and gcd(p,m)=1. Then, sG, the number of subgroups of order pk is congruent to 1modp. That is, sG=np+1, for some natural number, n.

Proof

Let Y:={YG||Y|=pk}. That is, Y is the set of all Sylow subgroups of G. Let G act on Y by conjugation, that is, g(Y):=g1Yg, which must also be a subgroup of order pk, and therefore also be in Y. We see that this action induces a permutation on Y.

Let YY be an element of Y. We know that Y has at least one element by Sylow’s first theorem, which we proved near the end of the previous entry (Counting Orbits, Sylow’s Theorems, etc…). Since Y is a subgroup of G, we can consider its actions on Y. As we’ve shown in previous entries, the action of the subgroup Y splits Y into disjoint orbits, and for any XY,

|OrbY(X)||StabY(X)|=|Y|=pk

This means that the number of elements in each orbit must be a power of p.

Notice that if we let X=Y, so we consider Y’s own orbit, it must have size equal to 1, because for any yY, y1Yy=Y. Now we prove that this only holds for X=Y.

Let XY be an arbitrary Sylow p-subgroup. Suppose that OrbY(X)={X}, then that means that YStabG(X)G. This holds because OrbX={X} means that every yY’s action is trivial, i.e. y1Xy=X. Therefore yStabG(X), the set of all elements of G whose action on X is trivial.

We also easily see that XStabG(X), because every xX will have action x1Xx=X.

Also, since the action we’re considering is conjugation, StabG(X)=NG(X), the normalizer of X in G. This enables us to apply the homomorphism lemmas we proved just before this proof. If we let

α:StabG(X)StabG(X)/X

be the natural homomorphism, then YX, which equals α1α(Y) is a subgroup of StabG(X), which has size equal to |Y||X||YX| which is equal to p2k|YX|. Notice that |YX| is a subgroup of Y (it is also a subgroup of X, but we only need to consider one group for this), that means that |YX| divides |Y|=pk, so |YX|=pj, for some jk. This means that the order of YX=p2kpj=p2kj.

YXNG(X)G, which means that YXG, and since |G|=pkm, we can conclude that p2kj|pk, since the order of a subgroup must divide the order of the group. Therefore, 2kjk, which means that jk. We also have YYX, so similarly, 2kjk, so jk. Therefore j=k. This means that |YX|=pk, which means that it is equal to Y and to X, because it is a subgroup of both of them, but it has exactly as many elements as both. Therefore, Y=X, as we wanted to show.

To summarize what we’ve shown about the orbits of Y, under the action of the subgroup Y. Y’s orbit is equal to just {Y}, while every other YXY has an orbit that has multiple subgroups in it. We also showed (using the orbit stablizer theorem) that pi=|OrbP(X)|, for all XY, therefore if we let OXl be the distinct orbits of Y, where the Xl’s range over orbit representatives, we see that |Y| is equal to Xl|OXl|, which is equal to XlpkXl, where only one OXl has size equal to 1, namely, OY. Therefore the sum is equal to 1+np, for some natural number n.

QED

Lemma (orbits and subgroup actions)

Let H, and K be two subgroups of G, such that HK. Let G act on a set S. Let Hs be an orbit under the action of the subgroup H containing sS. Then there exists an orbit Ks (where the s’s are the same) under the action of the subgroup K, such that HsKs.

What this means is that if G acts on a set S, and H is a subgroup of G which is also a subgroup of the subgroup K, then H’s orbits are sub-orbits of K’s orbits.

Proof

Let s1 and s2 belong to the same orbit under the action of the subgroup H. Then that means that there exists some hH, such that h(s1)=s2. Since HK, that same hK, so that h(s1)=s2 still holds under the action of the subgroup K. Therefore they’re still in the same orbit.

QED

Notice, however, that the orbits under the actions of the subgroup K may be larger. Theorem (Sylow’s third): All Sylow p-subgroups are conjugates. That is, for any two Sylow p-subgroups, X, and YG, there is a gG, such that g1Xg=Y.

Proof

This proof draws on the last proof quite a bit. However, instead of just considering the orbits of Y under the action of a subgroup Y, we consider consider the orbits of Y under two distinct Sylow p-subgroups as well as the orbits under the action of the entire group G. We have to be careful to remember what subgroup (even all of G) we’re dealing with.

To start, suppose that under the action of conjugation from the entire group G, there is more than one orbit of Y. This would mean that the theorem is false (we’re going for a contradiction). Let X1 and X2 be elements of two distinct orbits.

Now we switch from considering the orbits under the action of the entire group G, to considering the orbits under the action of conjugation by elements of X1 only. By the above lemma, the orbits under the action of the subgroup X1 will be sub-orbits of the orbits under the action of all of G.

In the proof of Sylow’s second theorem, we showed that OrbX1(X1)={X1}, and that all other orbits will have sizes that are a power of p, since X1 is a p-group. This means that OrbG(X1), the orbit of X1 under the action of all of G must have size equal to 1+pn1, for some natural number n1. We see this by considering that the larger orbit under the action of G can be broken up into disjoint sub-orbits under the action of X1. Each one of those orbits has a size that’s a power of p, but only OrbX1(X1) has only one element in it. Again, this is a very similar argument to what we made in the previous Sylow theorem.

Now consider how large the orbit OrbG(X2) is. This is the orbit of X2 under the action of all of G. By the same argument we made above, but with X2 substituted for X1, we see that it must have a size =1+pn2, for some natural number n2. However, when we consider the sub-orbits under the action of the subgroup X1, we see that every sub-orbit has a size which is divisible by p. This is because there isn’t an orbit of size 1. That only happens for the orbit of X1, but by our choices of X1 and X2, they belong to separate orbits of Y under the action of all of G. Therefore this orbit (under the action of all of G) must have size =pn3, for some natural number n3, but that’s impossible, because there is no natural number which is equal to 1+pn2 and pn3.

To see why, suppose a=1+pn2, and a=pn3. Then 1=apn2=pn3pn2, which implies that 1=p(n2n3), which means that p divides 1, which cannot happen.

Therefore, the assumption that there were more than one G orbits of Y is false, which proves the theorem.

QED

Corollary (another restriction on the number of Sylow p-subgroups):

Let sG denote the number of Sylow p-subgroups of a group G. Then sG|m, where |G|=pkm, and gcd(p,m)=1.

There are situations where this can allow us to prove that a Sylow p-subgroup is normal, for example.

Proof

sG=|orbG(X)|, where X is any Sylow p-subgroup (by the above theorem). By the orbit-stabilizer theorem, we have sG|StabG(X)=|G|=pkm. But, by the second Sylow theorem, sG=1+pn, so sGp, which means that sG|m.

QED

Finally we prove a result that doesn’t have anything to do with the Sylow theorems. It’s just a nice little result.

Theorem (counting orbits by counting fixes)

Let G be a group acting on a set X. I.e. G is a subgroup of SX, the group of all permutations on X. Then the number of orbits of X induced by the actions of G is equal to 1|G|gG|fix(g)|.

Proof

The first and more important step is to determine the size of

F:={(g,x)|gG,xX,where g(x)=x}

Notice that for a given xX, x will be in a pair for every gG, such that g(x)=x. That is just the set StabG(x). This means that

|F|=xX|StabG(x)|

On the other hand, for any given gG, g will be in a pair for every x, such that g(x)=x, which is just the set fix(g). Therefore

F|=gG|fix(g)|

Combining them, we get

xX|StabG(x)|(1)=gG|fix(g)|

We also have

xX|StabG(x)|=orbiti(xorbiti|StabG(x)|)

And by the orbit-stabilizer theorem, the right hand side is equal to

orbiti|G|=(2) # of orbits |G|

Therefore, if we combine (2) with (1) and divide by |G|, we get the desired equation.

QED