S. D. V. Stephens

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

26 Semidirect Products and First Applications of Sylow

· Denis Auroux

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

The Sylow theorems

If and , then in general there is no reason for to contain an element of order , or even a subgroup of order : the “converse to Lagrange’s theorem” fails.

Example 26.1. (resp. ) has no subgroup of order 6 (resp. 30): such a subgroup would be normal.

Fix a prime (which divides ) and write , .

Definition 26.2. A subgroup of order is called a Sylow -subgroup of .

Theorem 26.3 (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, ).

Example 26.4 (Classify groups of order 15).

Show solution

If , then there exist Sylow subgroups with , . The numbers of such Sylow subgroups: This implies and are normal! (Since their conjugates , are also Sylow subgroups, but and are the unique such.) Using a criterion for direct products coming up next, this implies Every group of order 15 is cyclic!

Digression: normal subgroups, semidirect products and direct products

Let’s say is a normal subgroup; then we have an exact sequence This does not imply that , or in fact even that contains a subgroup isomorphic to !

Example 26.5. is a subgroup, , but has no subgroup .

On the other hand, assume we can identify with a subgroup of which contains one element of each coset of , so that the inclusion satisfies . This means: and are subgroups of , is normal, and every coset of (left = right cosets since is normal) contains a unique element of : Thus, every element of can be uniquely expressed as , , . So we have a bijection of sets This need not be a group isomorphism! (In particular because need not be a normal subgroup of .) However, since is normal, we do know that ; in fact This can be interpreted as a semidirect product of and :

Definition 26.6. Given groups and , and an action of on by automorphisms, i.e. a homomorphism , we define the semidirect product :

  • as a set: ;
  • group law: .

(Check: this satisfies the group axioms; in particular it’s associative.)

In the above setting, acts on the normal subgroup by conjugation: , and then we find that . To summarize:

Proposition 26.7. If and are subgroups of , normal, such that every coset of contains a unique element of ( every element of is uniquely ), then is isomorphic to a semidirect product .

Example 26.8. , where ( a 3-cycle) is the alternating subgroup (normal). We can realize as the subgroup , a transposition (not normal). So , where acts on by conjugation: .

Similarly, , with the rotations and , so .

These are not direct products.

Remark. If is finite, , and , then every coset of contains a unique element of ; so, assuming normal, we have a semidirect product, by the proposition.

Indeed: the homomorphism , (the composition ) has , so it is injective, and , so it is bijective.

Alternatively: if , then , so and . Thus the products , , , are all distinct; every element of has at most one such expression, so exactly one since .

Finally:

Proposition 26.9. If both and are normal subgroups of , and every element of can be uniquely expressed as , , ( every coset of one subgroup contains a unique element of the other subgroup), then (i.e. the semidirect product is actually a direct product).

Proof. This is because cosets intersect in a single element: (because , and left multiplication by gives this). And, since and are normal, so , hence , showing that , , is now a group isomorphism. ∎

Corollary 26.10. If is finite, are normal subgroups, and , then .

Remark. The condition is e.g. automatic if (since is a subgroup of and , so its order divides and ).

Groups of order 15 and 21

So, returning to a group of order 15: the Sylow theorems imply has unique subgroups and of orders 3 and 5, which are normal (uniqueness , ). Since and , the criterion holds and so

Example 26.11 (Groups of order 21).

Show solution

Sylow gives the existence of subgroups of order 3 and of order 7. Also, the number of conjugate subgroups of each of these: and , so ; and , so could be either 1 or 7.

If , then and are normal (since they are equal to their conjugates), and the above criterion implies that Otherwise, if , then is normal but isn’t: we have a semidirect product . Let be a generator of and a generator of : then , and every element of is uniquely expressible as , , . What we need to know, to determine the group structure, is the expression of . Since is normal, , so for some , i.e. . This determines the group law.

Further investigation shows that in fact there exists a unique nonabelian group of order 21 up to isomorphism. The best way to prove existence is to construct it explicitly, e.g. as a subgroup of or of something else. This is on the homework!

Two comments on the Sylow theorems

Next time, we’ll look at the proof of the Sylow theorems. For now, a couple of comments.

  1. Recall: for all , the order of divides ; but the converse does not hold: in general, there exists of order . A corollary of Sylow’s first theorem (existence of Sylow -subgroups) is that the converse does hold for primes.

Corollary 26.12. 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 . ∎

  1. 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 .