sylow

CHATER FIVE

SYLOW THEOREMS

In this chapter, we will study CONJUGATE, NORMALIZER , P- . GROUPS, CENTRALIZER and SYLOW THEOREMS .
For any subgroup (H,*) of the group (G,*) and any element a? G, . we will study the following subset of G :
a* H * a-1 = { a * h * a-1 | h ? H }.

Theorem 5 – 1 .
If ( H , *) is a subgroup of the group (G, *) and a? G,
then ( a* H * a -1 , *) is subgroup of (G , * ).
Proof . H .W.

Definition 5 - 1.
Let (H,*) be subgroup of the group (G,*) and a ? G. The . subgroup (a*H*a-1 ,*) is called the CONJUGATE SUBGROUP . of H induced by the element a.

Example 1 .
Consider (S3,o), if H1 and H2 are subgroups of the S3 such that . o(H1) ? o(H2 ). Find a*H1*a-1 and a*H2 *a-1 for all a in G.


Definition 5 - 2 .

Let (H,*) be subgroup of the group (G,*) , if a*H*a-1 = H for . some a ? G , then we say that the subgroup H is INVARIANT or
SELF - CONJEGAT under a .
Example 2 .

Are H1 and H2 in example 1 invariant?

NOTE:

Any subgroup is invariant under each of its own elements .

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Theorem 5 - 2 .

A subgroup (H,*) is normal in (G,*) if and only if (H,*) is . invariant under each element of G .
Proof :
H – W - .
Exercis .

Prove that conjugate subgroups are isomorphic.


Now ,if (H,*)and (K,*) are subgroups of the group (G,*) we will . study another concept , this concept is the set of elements of K . under which H is invariant ;in other words, the set

NK(H) = {k ? K | k* H *k-1= H }.

Example 3 .

In example 1 find NK(H) when H=H1 and K=H2.

Theorem 5 - 3 .

If (H , *) and (K , *) are subgroups of the group (G , *) ,then . (NK(H),*) is subgroup.
Proof:
H – W .

Definition 5 - 3 .
Let ( H ,*) and (K ,*) be subgroups of the group (G ,*) , NK (H) is . called the NORMALIZER subgroup of H in K.

NOTES:

1- If K=G, we write N(H) instead of NG(H).
2- H is not necessarily normal in G.
3- H is normal in N (H) [N(H) is the largest subgroup of G in
which H is normal.


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We can now count the conjugate subgroups of subgroup by:

Theorem 5 - 4 .
The number of distinet conjugates of a subgroup (H ,*) of (G ,*) . induced by the elements of a subgroup (K,*) is equal to [K:NK (H)], . the index of NK (H) in K .
Proof:
If k1,k2?K, then
? k1 *H * k1-1 = k2 * H * k2-1
? H = (k1-1 * k2) * H *( k1-1 *k2 )-1
? k1-1 * k2 ? NK (H)
? k1 * NK(H) = k2 * NK (H)
Hence , the number of distinet conjugate subgroups of H by
elements of K is equal to the number of distinet cosets of
NK (H) in K ,
that is mean , it is equal to [ K : NK (H) ] .

Corollary .
Let (H,*) and (K,*) be subgroups of the group (G,*) . If
H is invariant under n elements of K , then H has o(K) / n
Conjugate subgroups by elements of K