PECULIARITIES OF SILOV'S THEOREMS AND THEIR APPLICATIONS

ABSTRACT

The article is devoted to the consideration of Sylow's theorems and their proofs. Discussion of Sylow's theorems, the study of definitions, as well as their areas of application is the basis of the study.

The result of this discussion is the systematization of the accumulated material on this issue

Keywords: Sylow's theorems, Abelian groups, mathematics, Lagrange, Sylow subgroups.

In 1862, an event occurred in the world of mathematics and it cannot be overestimated – by chance, Sylow replaced a professor of Galois Theory at the University of Christiania. As a result, by 1872, Sylow's theorems were created, which were an important result of the whole life of this mathematician, and also a striking result of finite group theory in the direction of a partial reversal of the Lagrange theorem.

First Sylow Theorem.

For every prime factor p of the order of a finite group G, there exists a Sylow p-subgroup of S ⊂ G.

Proof:

1. If |G| = p k, then S = G.

2. If G – abelian group, then S = Torp(G). For the general case, we will carry out the proof by induction in order G.

Proposition 1.

In it there will be a certain subgroup A such that |A| = p. It is clear that A is a normal subgroup in G. In this case, |G/A| = n p, where n = |G|. By the assumption of induction in G/A there is a Sylow p-subgroup B in G/A. Consider π -1 A (B) ⊂ G. We have, |π -1 A (B)| = |Ker(πa | π −1 A (B) )| · |Im (πa | π −1 A (B) )| = |A| · |B| = p k . We can take S = π -1 A (B).

Proposition 2.

|Z(G)| is not divisible by p. Let the decomposition of a group G on classes of conjugate elements. We know, that the elements of the center are called conjugacy classes, which consist of a single element. We can see, that |G| divided by p and we can find conjugacy classes C, which |C|6= 1 not divided by p. Let g ∈ C. Let |Cent(g)| = |G| |C| < |G|. С another side, |Cent(g)| divided by p k . On the assumption of induction, we have Sylow subgroup S ⊂ Cent(g) ⊂ G and |S| = p k[5].

If the Sylow subgroup is unique, then it is normal.

Proof:

Let gSg −1 is Sylow subgroups G. But |gSg −1 | = |S|. It’s clear that |gSg −1 | ≤ |S|, another side, S = g −1 (gSg −1 )g, mean, S ≤ |gSg −1 . We have, gSg −1 – there is a Sylow subgroup G, а mean, gSg −1 = S, S is normal.

Second Sylow Theorem.

1. Any p-subgroup of G is contained in some Sylow.

2. The Sylow p-subgroups of a group G are conjugate.

Proof:

When m = 1 it is clear. Let m > 1. Let S is Sylow subgroup, |S| = p k. Let H ⊂ G is subgroup of order p l , l ≤ k. Consider the action of H on the set of left adjacent classes in S: h · gS = (hg)S. Is it clear, if gS = g 0S, then g 0 = gs and for someone s ∈ S. So, hg0 = hgs and hgS = hg0S.

From Lagrange's theorem, the number of left adjacent classes in S is equal to |G||S|= m [2]. We have, |H| = p l = |St(gS)| · |Orb(gS)|, so the order of each orbit is either 1 or a power of p. Since the sum of the orders of orbits not divided by p, there is an orbit of one element, that is, gs = gS. Hence g −1hg ∈ S, that is, h ∈ gSg -1 .

We can conclude, that H ⊂ gSg −1, где |gSg −1 | = p k . If |H| = p k , then H = gSg −1 . So, we can conclude that, The Sylow theorems assert a partial converse to Lagrange's theorem.

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