Page 96 - Ellingham, Mark, Mariusz Meszka, Primož Moravec, Enes Pasalic, 2014. 2014 PhD Summer School in Discrete Mathematics. Koper: University of Primorska Press. Famnit Lectures, 3.
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3.3 Finite simple groups
Orbit-Stabilizer Theorem, |H \G | is a power of p .
Faithful actions and Iwasawa’s Lemma
From here on we consider only faithful actions. We say that such an action of G on X is
regular if it is transitive and the point stabilizer is trivial. From the above we see that a
regular action of G is isomorphic to the action of G on itself by right multiplication.
Let G act faithfully on X and let N be a normal subgroup of G whose action on X
is regular. Then we can identify X with N , so that N acts by right multiplication. To
be more precise, choose x ∈ X and observe there is a bijection between N and X under
which n ∈ N corresponds to x n ∈ X . Under the above bijection, the action of stabG (x )
on N by conjugation corresponds to the given action on X . To see this, take g ∈ stabG (x )
and suppose that y g = z . Let h, k ∈ N correspond to y , z ∈ X under the above bijection,
that is, x h = y , x k = z . Then x (g −1h g ) = x h g = y g = z . Since the action is faithful, we
conclude that g −1h g = k , as required.
Theorem 3.3.8 (Iwasawa’s Lemma) Let G be a group with a faithful primitive action on
X . Suppose there exists an abelian normal subgroup A of stabG (x ) with the property that
the conjugates of A generate G . Then any non-trivial normal subgroup of G contains G .
In particular, if G is perfect, then it is simple.
PROOF. Let N be a non-trivial normal subgroup of G . By Proposition 3.3.5, N acts transi-
tively on X , therefore N ≤ stabG (x ). By Proposition 3.3.4, stabG (x ) is a maximal subgroup
of G . Hence N stabG (x ) = G . Take g ∈ G and write it as g = n h, where n ∈ N and
h ∈ stabG (x ). Then g A g −1 = n hAh−1n −1 = n An −1. We conclude that g A g −1 ≤ N A. By
our assumption it follows that G = N A. Now, G /N =∼ A/(A∩N ) is abelian, hence G ≤ N .
3.3.2 Symmetric groups and alternating groups
Here we examine the normal subgroups of Sn and prove that if n ≥ 5, then the alternating
group An is simple.
Proposition 3.3.9 Two elements of Sn are conjugate if and only if they have the same cycle
structure.
PROOF. If π ∈ Sn and γ = (a 1 a 2 ... a k ) is a cycle, then γπ = (a π a π ... a π ).
1 2 k
Proposition 3.3.10 The alternating group An is generated by the 3-cycles.
Orbit-Stabilizer Theorem, |H \G | is a power of p .
Faithful actions and Iwasawa’s Lemma
From here on we consider only faithful actions. We say that such an action of G on X is
regular if it is transitive and the point stabilizer is trivial. From the above we see that a
regular action of G is isomorphic to the action of G on itself by right multiplication.
Let G act faithfully on X and let N be a normal subgroup of G whose action on X
is regular. Then we can identify X with N , so that N acts by right multiplication. To
be more precise, choose x ∈ X and observe there is a bijection between N and X under
which n ∈ N corresponds to x n ∈ X . Under the above bijection, the action of stabG (x )
on N by conjugation corresponds to the given action on X . To see this, take g ∈ stabG (x )
and suppose that y g = z . Let h, k ∈ N correspond to y , z ∈ X under the above bijection,
that is, x h = y , x k = z . Then x (g −1h g ) = x h g = y g = z . Since the action is faithful, we
conclude that g −1h g = k , as required.
Theorem 3.3.8 (Iwasawa’s Lemma) Let G be a group with a faithful primitive action on
X . Suppose there exists an abelian normal subgroup A of stabG (x ) with the property that
the conjugates of A generate G . Then any non-trivial normal subgroup of G contains G .
In particular, if G is perfect, then it is simple.
PROOF. Let N be a non-trivial normal subgroup of G . By Proposition 3.3.5, N acts transi-
tively on X , therefore N ≤ stabG (x ). By Proposition 3.3.4, stabG (x ) is a maximal subgroup
of G . Hence N stabG (x ) = G . Take g ∈ G and write it as g = n h, where n ∈ N and
h ∈ stabG (x ). Then g A g −1 = n hAh−1n −1 = n An −1. We conclude that g A g −1 ≤ N A. By
our assumption it follows that G = N A. Now, G /N =∼ A/(A∩N ) is abelian, hence G ≤ N .
3.3.2 Symmetric groups and alternating groups
Here we examine the normal subgroups of Sn and prove that if n ≥ 5, then the alternating
group An is simple.
Proposition 3.3.9 Two elements of Sn are conjugate if and only if they have the same cycle
structure.
PROOF. If π ∈ Sn and γ = (a 1 a 2 ... a k ) is a cycle, then γπ = (a π a π ... a π ).
1 2 k
Proposition 3.3.10 The alternating group An is generated by the 3-cycles.
