Actions and Sylow theorems · Definition
Sylow theorems
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.
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
- How many Sylow \(5\)-subgroups can a group of order \(15\) have?
- In a group of order \(12\), which is a possible value of \(n_3\)?
- What is the order of a Sylow \(2\)-subgroup of a group of order \(72\)?
- A group of order \(56\) has \(n_7\) equal to:
- A group \(G\) of order \(20\) has a unique Sylow \(5\)-subgroup. How many elements of order \(5\) does \(G\) have?
- A group of order \(21\) has \(n_7=1\). What follows?
- How many Sylow \(2\)-subgroups does the symmetric group \(S_3\) have?
- Every group of order \(35\) is:
- If \(P\) is a Sylow \(3\)-subgroup of \(G\) and \(g\in G\), the conjugate \(gPg^{-1}\) is:
- A group of order \(12\) has \(n_3=4\). How many elements of order \(3\) does it have?
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