6 1. Sylow Theory
1.5. Corollary. Let x G G, where G is a finite group, and let K be the
conjugacy class of G containing x. Then \K\ = \G :
Proof. The class of x is the orbit of x under the conjugation action of G on
itself, and the stabilizer of x in this action is the centralizer
|if | = \G :
C G ( # ) | ,
as required.
1.6. Corollary. Let H C G be a subgroup, where G is finite. Then the total
number of distinct conjugates of H in G, counting H itself, is \G :
Proof. The conjugates of H form an orbit under the conjugation action of
G on the set of subsets of G. The normalizer NQ(H) is the stabilizer of H
in this action, and thus the orbit size is \G :
as wanted.
Problems 1A
1A.1. Let H be a subgroup of prime index p in the finite group G, and
suppose that no prime smaller than p divides |G|. Prove that H G.
1A.2. Given subgroups H,K C G and an element ^ G G , the set HgK
{hgk \ h E H, k G K} is called an (ii, K)-double coset. In the case where
H and K are finite, show that \HgK\ = \H\\K\/\K n H\.
Hint. Observe that HgK is a union of right cosets of H, and that these
cosets form an orbit under the action of K.
Note. If we take g = 1 in this problem, the result is the familiar formula
\HK\ = \H\\K\/\HnK\.
1 A.3. Suppose that G is finite and that H^ K C G are subgroups.
(a) Show that \H \ H C\ K\ \G \ K\, with equality if and only if
HK = G.
(b) If \G : ii| and \G : K\ are coprime, show that HK G.
Note. Proofs of these useful facts appear in the appendix, but we suggest
that readers try to find their own arguments. Also, recall that the product
HK of subgroups H and K is not always a subgroup. In fact, HK is a
subgroup if and only if HK KH. (This too is proved in the appendix.)
If HK = KH, we say that H and K are permutable.
1A.4. Suppose that G HK, where H and K are subgroups. Show that
also G =
for all elements x,y G G. Deduce that if G
for a
subgroup H and an element x G G, then H G.
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