What is Sylow's theorem and how does it relate to finite groups?

  • Thread starter Thread starter halvizo1031
  • Start date Start date
  • Tags Tags
    Theorem
Click For Summary
SUMMARY

Sylow's theorem states that for a finite group G whose order is divisible by a prime p, there exists at least one subgroup of G with order equal to the highest power of p dividing the order of G. Specifically, if p^M is the highest power of p that divides |G|, then G contains a subgroup of order p^M. The proof involves using group actions and the orbit-stabilizer theorem, demonstrating that the size of certain subsets must yield a subgroup of the desired order.

PREREQUISITES
  • Understanding of finite group theory
  • Familiarity with group actions and the orbit-stabilizer theorem
  • Knowledge of binomial coefficients and their properties
  • Basic concepts of subgroup orders and divisibility
NEXT STEPS
  • Study the proof of Sylow's theorem in detail
  • Explore examples of finite groups and their Sylow subgroups
  • Learn about the applications of Sylow's theorem in group classification
  • Investigate related concepts such as normal subgroups and group homomorphisms
USEFUL FOR

Mathematics students, particularly those studying abstract algebra, group theorists, and educators seeking to deepen their understanding of finite groups and Sylow's theorem.

halvizo1031
Messages
77
Reaction score
0

Homework Statement


LET G BE A FINITE GROUP WHOSE ORDER IS DIVISIBLE BY THE PRIME P. SUPPOSE P^M IS THE HIGHEST POWER OF P WHICH IS A FACTOR OF |G|AND SET K=(|G|/P^M), THEN THE GROUP G CONTAINS AT LEAST ONE SUBGROUP OF |P^M|.

I have the proof but can someone explain it in simpler terms? Maybe even a few examples please.



Homework Equations





The Attempt at a Solution


Let X denote the collect of all subsets of G which have p^m elements and let G act on X by left translation so that the group element g is in G sends the subset A in X to gA. The size of X is the binomial coefficient kp^m choose p^m which is not divisible by p. Hence, there must be an orbit G(A) whose size is not a multiple of p. We have |G|=|G(A)|*|G(subA)|, consequently |G(subA)| is divisible by p^m. Now G(subA) is the stabilizer of A, so if a is in A and g is in G(subA), the ga is in A. This means that the whole right coset of G(subA)a is contained in A whenever a is in A and |G(subA)| cannot exceed p^m. Therefore, G(subA) is a subgroup of G which has order p^m.
 
Last edited:
Physics news on Phys.org
The statement is pretty simple. It says that if you have a Group of order n, where n has factoriaztion p^k * m, where p is a prime number, then there exists a subgroup of order p^k.

The proof is a bit tricky, since it's in direct. What about it is confusing you?
 
could you maybe give me a few examples? It makes sense but the proof is a bit rough for me.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
5
Views
2K