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|aH\cap Hb| and |H|, here H\leq G

Source: 4-th Hungary-Israel Binational Mathematical Competition 1993

May 27, 2007

group theoryabstract algebrasuperior algebrasuperior algebra unsolved

Problem Statement

In the questions below:

G

is a ﬁnite group;

H \leq G

a subgroup of

G; |G : H |

the index of

H

in

G; |X |

the number of elements of

X \subseteq G; Z (G)

the center of

G; G'

the commutator subgroup of

G; N_{G}(H )

the normalizer of

H

in

G; C_{G}(H )

the centralizer of

H

in

G

; and

S_{n}

the

n

-th symmetric group. Let

H \leq G

and

a, b \in G.

Prove that

|aH \cap Hb|

is either zero or a divisor of

|H |.

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