Page 97 - Ellingham, Mark, Mariusz Meszka, Primož Moravec, Enes Pasalic, 2014. 2014 PhD Summer School in Discrete Mathematics. Koper: University of Primorska Press. Famnit Lectures, 3.
P. 97
mož Moravec: Some Topics in the Theory of Finite Groups 85
PROOF. Note that 3-cycles are even permutations. If π is any even permutation, then it
can be written as a product of an even number of transpositions. Thus we only need to
consider products of two transpositions. If a ,b, c , d ∈ {1, 2, . . . , n} are pairwise different,
then the following clearly hold:
(a b )(a b ) = 1,
(a b )(a c ) = (a b c ),
(a b )(c d ) = (a b c )(a d c ),
and we are done.
Proposition 3.3.11 The following are equivalent for π ∈ An :
1. The Sn conjugacy class of π splits into two An -conjugacy classes;
2. There is no odd permutation which commutes with π;
3. π has no cycles of even length, and all of its cycless have distinct lengths.
PROOF. Let us proove that (1) is equivalent to (2). The group Sn acts transitively on An
by conjugation. We have that CAn (π) = CSn (π) ∩ An . If (2) holds, then CAn (π) = CSn (π),
therefore π has |An : CAn (π)| = |Sn : CSn |/2 conjugates in An . Thus (1) follows. If (2) does
not hold then |CAn (π)| = |CSn (π)|/2, and π has |An : CAn (π)| = |Sn : CSn | conjugates in An .
Therefore (1) does not hold.
Now we prove that (2) and (3) are equivalent. If π has a cycle of even length, then this
cycle is an odd permutation commuting with π. If π has only cycles of odd length, and
two cycles of the same length , then a permutation interchanging them is a product of
transpositions commuting with π. This proves that (2) implies (3). Assume now that (3)
holds. Then any permutation commuting with π fixes each of its cycles and acts on it as
a power of the corresponding cycle of π, hence it is an even permutation.
Proposition 3.3.12 The group A5 is simple.
PROOF. A lazy proof is
gap> IsSimple( AlternatingGroup( 5 ) );
true
A formal proof goes as follows. The conjugacy classes of A5 can be determined using
Proposition 3.3.11:
• Representative (∗)(∗)(∗)(∗)(∗): this class has size 1 and does not split into two con-
jugacy classes of A5;
PROOF. Note that 3-cycles are even permutations. If π is any even permutation, then it
can be written as a product of an even number of transpositions. Thus we only need to
consider products of two transpositions. If a ,b, c , d ∈ {1, 2, . . . , n} are pairwise different,
then the following clearly hold:
(a b )(a b ) = 1,
(a b )(a c ) = (a b c ),
(a b )(c d ) = (a b c )(a d c ),
and we are done.
Proposition 3.3.11 The following are equivalent for π ∈ An :
1. The Sn conjugacy class of π splits into two An -conjugacy classes;
2. There is no odd permutation which commutes with π;
3. π has no cycles of even length, and all of its cycless have distinct lengths.
PROOF. Let us proove that (1) is equivalent to (2). The group Sn acts transitively on An
by conjugation. We have that CAn (π) = CSn (π) ∩ An . If (2) holds, then CAn (π) = CSn (π),
therefore π has |An : CAn (π)| = |Sn : CSn |/2 conjugates in An . Thus (1) follows. If (2) does
not hold then |CAn (π)| = |CSn (π)|/2, and π has |An : CAn (π)| = |Sn : CSn | conjugates in An .
Therefore (1) does not hold.
Now we prove that (2) and (3) are equivalent. If π has a cycle of even length, then this
cycle is an odd permutation commuting with π. If π has only cycles of odd length, and
two cycles of the same length , then a permutation interchanging them is a product of
transpositions commuting with π. This proves that (2) implies (3). Assume now that (3)
holds. Then any permutation commuting with π fixes each of its cycles and acts on it as
a power of the corresponding cycle of π, hence it is an even permutation.
Proposition 3.3.12 The group A5 is simple.
PROOF. A lazy proof is
gap> IsSimple( AlternatingGroup( 5 ) );
true
A formal proof goes as follows. The conjugacy classes of A5 can be determined using
Proposition 3.3.11:
• Representative (∗)(∗)(∗)(∗)(∗): this class has size 1 and does not split into two con-
jugacy classes of A5;
