S. D. V. Stephens

notes / Studies in Algebra and Group Theory / chapter 7: Finite Groups

28 Proofs of the Sylow Theorems; Groups of Order 12

· Denis Auroux

Original Professor Auroux’s handwritten notes (PDF), behind a password for now:

The Sylow theorems

Fix a prime (which divides ) and write , .

Definition 28.1. A subgroup of order is called a Sylow -subgroup of .

Theorem 28.2 (Sylow theorems (Sylow, 1872)).

  1. For every prime , a Sylow -subgroup of exists.
  2. All Sylow -subgroups are conjugates of each other: if are Sylow -subgroups, then there exists such that . Moreover, any subgroup with a power of is contained in a Sylow -subgroup.
  3. Let be the number of Sylow -subgroups of . Then , and (or equivalently, ).

We saw last time: if , then the unique Sylow -subgroup is a normal subgroup.

Example 28.3.

  • contains exactly one subgroup of order 3 and one of order 5, both are normal, and .
  • there is a unique subgroup of order 7 (normal), and either or is a semidirect product of and .

For a -group (), Sylow tells us exactly nothing! Namely, a Sylow -subgroup has elements, and the only such is itself. Thus, in the Sylow approach to classification, -groups are the hardest to classify. In fact, the number of different -groups grows dramatically with the exponent ! E.g. for :

order number of groups
1 (cyclic)
2 (, )
5
14
51

… and already 56092 for .

A corollary of Sylow’s first theorem (existence of Sylow -subgroups):

Corollary 28.4. If and is prime, then contains an element of order .

Proof. Let be a Sylow -subgroup, and let such that . Since the order of divides , it is for some . Now has order . ∎

Proof of the first theorem

The first two theorems are proved by studying the action of on its subsets by left multiplication. The proof of Sylow’s first theorem uses two lemmas.

Lemma 28.5. Given with , we have .

Proof. The highest power of dividing or is exactly the highest power of dividing (look mod !); hence the numerator and denominator have the same powers of in their prime factorization, and the end result has no powers of . ∎

Lemma 28.6. Let be any subset, and consider the action of on by left multiplication. Then the stabilizer of , has dividing .

Proof. Let ; then acts on by left multiplication ( for all ), and so is a union of orbits for various . But each orbit is a (right) coset of , and has . Since is a union of such orbits, divides . ∎

Now we can give the proof of Sylow’s first theorem (existence of Sylow subgroups).

Proof of Sylow’s first theorem. Let : all subsets of with elements. Consider the action of on by left multiplication, , and partition into orbits for this action. By Lemma 1, , so there exists an orbit such that . Since divides , we find that .

But by Lemma 2, divides . So . We’re done: is a Sylow -subgroup! (And in fact was a right coset of .) ∎

Proof of the second theorem

Next we prove Sylow’s second theorem, formulated as:

Theorem 28.7 (Sylow’s second theorem). If is a Sylow -subgroup and is any -subgroup, then there exists a conjugate with .

(For , this says all Sylow -subgroups are conjugate.)

Proof. Let be the set of left cosets of ; then acts on (by left multiplication) transitively (i.e. there is only one orbit); ; and there exists , namely itself, such that . (Any -action on a set with these properties would work just as well.)

Now restrict the action of on to a -subgroup . The -action on has orbits of size dividing , hence a power of . Since , there is at least one fixed point (i.e. there exists with for all ). Thus , which is conjugate to since are in the same orbit of .

(Concretely: assume the coset is fixed by , i.e. for all . Then for all , , so , hence . Thus .) ∎

Normalizers and conjugate subgroups

Before we can prove the 3rd theorem, we need to discuss normalizers and conjugate subgroups.

Question 28.8. Given a group and a subgroup , what is the largest subgroup such that is normal inside ?

Observe: the issue is whether . This might not hold for all , but needs to hold for all .

Definition 28.9. The normalizer of a subgroup is This is a subgroup of , and for subgroups , is normal in iff .

Example 28.10. :

  • ( is normal in !), even though, for a transposition, ; still ✓.
  • for a transposition . (Note: .)

The normalizer measures how close is to being normal in : if it is, then .

acts by conjugation on the set of all of its subgroups. The orbit of is the set of its conjugate subgroups . (If is normal, then .) The stabilizer of is . So by orbit-stabilizer, (and ).

Proposition 28.11. The number of subgroups conjugate to in is .

Proof of the third theorem

Now the proof of Sylow’s third theorem (the number of Sylow -subgroups satisfies and ).

Proof of Sylow’s third theorem. Consider the action of on the set of Sylow -subgroups by conjugation. By the 2nd theorem, this action is transitive (all Sylow -subgroups are conjugate), and if is any Sylow -subgroup, the stabilizer is (the normalizer), and so Since are subgroups and , , and hence Next, we restrict to the conjugation action on the set of all Sylow -subgroups, and observe that itself is fixed ( for all ), so this gives an orbit of size 1. We claim it’s the only one.

Indeed, let be a Sylow -subgroup of such that for all (orbit ). This means . But is so and are Sylow -subgroups of ! By Sylow’s 2nd theorem, they’re conjugate subgroups of . However, is normal in (by definition!). Therefore . This shows the only orbit of size 1 for the action of by conjugation on the set of Sylow -subgroups of is itself.

Since the size of an orbit of an -action divides , all other orbits have size divisible by . We conclude that ∎

Groups of order 12

One more example, to show that things can get more complicated quickly.

Example 28.12 (Classify groups of order 12).

Show solution

If , then Sylow gives

  • a subgroup , ; the number of these is (, );
  • a subgroup , ; the number is (, ).

At least one of these is normal. Indeed, if , then the nontrivial elements of all have order 3, and (its order divides 3 and is ), so we have 8 elements of order 3. So there are at most 4 elements of order in , hence and is normal.

If both and are normal, then (using , ; see last time), and so is abelian, one of

If is normal but isn’t, consider the action of on by conjugation. Conjugation by a nontrivial element of maps to itself, but doesn’t fix any of the 3 others: indeed, recall the stabilizer of is , and by orbit-stabilizer, so . So a nontrivial element of acts on by a 3-cycle permuting , and similarly for the others. Hence the action of on gives a homomorphism , sending the generators of the to 3-cycles. This implies , hence , and ( the semidirect product ).

If is normal but isn’t, then there are 2 subcases: or !

  • If : let be a generator, and let . Then is determined by the conjugation action of on , i.e. we need to know . We can’t have () or ( and commute, abelian). So instead . Then is generated by , with This group is unfamiliar to us: it is the semidirect product , where acts on the normal subgroup by
  • If : look at the conjugation action . Necessarily ; denote by its generator, let be such that generate , and let be a generator of . Then is generated by with We can check this is actually . (The subgroup generated by and is and normal in ; take rotation by , rotation by , any reflection.)

Thus there are 5 isomorphism classes of groups of order 12: