Actions and Sylow theorems · Definition

Sylow theorems

modern algebragroups

Before you start you need to know

The Sylow theorems answer the converse of Lagrange: if a prime power divides the order of the group, is there a subgroup of that size?

Definition

A finite group \(G\) of order \(p^k m\), with \(p\) prime and \(p\nmid m\), has subgroups of order \(p^k\), all conjugate to each other; their number \(n_p\) is restricted.

\[n_p\equiv1\pmod p,\qquad n_p\mid m\]

A Sylow \(p\)-subgroup is a subgroup of order \(p^k\), the largest power of \(p\) dividing \(|G|\). The theorems say three things: at least one exists; any two are conjugate, that is \(Q=gPg^{-1}\) for some \(g\); and their number \(n_p\) divides \(m\) and is congruent to \(1\) modulo \(p\). Moreover, \(n_p=[G:N_G(P)]\), the index of the normalizer of any of them.

The proof uses what we saw about actions. Existence comes from counting with an action of \(G\) on a family of subsets and from the fact that a \(p\)-group acting on a set has as many fixed points as its cardinality modulo \(p\). Conjugacy between Sylow subgroups and the formula for \(n_p\) come from letting \(G\) act by conjugation on the set of them and applying the orbit-stabilizer theorem.

The most used consequence is that \(n_p=1\) is equivalent to the Sylow \(p\)-subgroup being normal, since its conjugates coincide with it. To decide, list the divisors of \(m\) congruent to \(1\) modulo \(p\). Guiding example: in order \(15\) only \(n_3=n_5=1\) is possible, so both subgroups are normal and the group is \(\mathbb{Z}_{15}\); in order \(12\), \(n_3\in\{1,4\}\).

Typical mistakes: forgetting that the order of a Sylow subgroup is the maximal power and not any power, confusing «divides \(m\)» with «divides \(|G|\)», and believing that \(n_p=1\) always. These theorems show that many orders admit no simple groups, and they prepare the classification of small finite groups.

Think about it

Can there be a simple group of order \(15\)?

No. Since \(n_5=1\), the Sylow \(5\)-subgroup is a proper normal subgroup, so the group is not simple.

Exercises 10

References

  • David S. Dummit, Richard M. Foote, Abstract Algebra, 3rd edition, Wiley, 2004. §4.5
  • John B. Fraleigh, A First Course in Abstract Algebra, 7th edition, Pearson, 2002. §36
  • Joseph A. Gallian, Contemporary Abstract Algebra, 9th edition, Cengage, 2017. cap. 24