Abelian normal subgroup of core-free maximal subgroup is contranormal implies derived subgroup of whole group is monolith: Difference between revisions

Latest revision as of 19:27, 20 November 2022

Statement

Suppose $G$ is a Primitive group (?), $M$ is a core-free maximal subgroup of $G$ (or, the stabilizer of a point for a faithful primitive group action of $G$ ), and $A$ is an abelian normal subgroup of $M$ that is a contranormal subgroup of $G$ : $G$ is generated by the conjugates of $A$ in it. Then, the derived subgroup $[G, G]$ is the monolith of $G$ , i.e., every nontrivial normal subgroup of $G$ contains the commutator subgroup of $G$ .

Facts used

Proof

Given: A group $G$ , a core-free maximal subgroup $M$ of $G$ . An abelian normal subgroup $A$ of $M$ that is contranormal in $G$ . A nontrivial normal subgroup $N$ of $G$ .

To prove: $G / N$ is abelian.

Proof:

$M N = G$ : Since $M$ is core-free and $N$ is nontrivial normal, $M$ does not contain $N$ . Since $M$ is maximal, $M N = G$ .
$A N$ is normal in $G$ : Clearly, $N \leq A N \leq N_{G} (A N)$ . Also, $A$ is normal in $M$ and $N$ is normal in $G$ , so $M \leq N_{G} (A N)$ . Thus, $M N \leq N_{G} (A N)$ , so $N_{G} (A N) = G$ by step (1).
$A N = G$ : By assumption, $A$ is contranormal. Thus, the only normal subgroup of $G$ containing $A$ is $G$ . So, by step (2), $A N$ is normal in $G$ .
$G / N ≅ A / (A \cap N)$ is abelian: This follows from facts (1) and (2).

References

Textbook references

Book:Cohn, Page 120, ^{More info}

@@ Line 1: / Line 1: @@
 ==Statement==
-Suppose <math>G</math> is a [[fact about::primitive group]], <math>M</math> is a [[fact about::core-free maximal subgroup]] of <math>G</math> (or, the stabilizer of a point for a faithful primitive group action of <math>G</math>), and <math>A</math> is an [[fact about::abelian normal subgroup]] of <math>M</math> that is a [[fact about::contranormal subgroup]] of <math>G</math>: <math>G</math> is generated by the conjugates of <math>A</math> in it. Then, the [[fact about::commutator subgroup]] <math>[G,G]</math> is the [[monolith]] of <math>G</math>, i.e., every nontrivial [[normal subgroup]] of <math>G</math> contains the commutator subgroup of <math>G</math>.
+Suppose <math>G</math> is a [[fact about::primitive group]], <math>M</math> is a [[fact about::core-free maximal subgroup;3| ]][[core-free maximal subgroup]] of <math>G</math> (or, the stabilizer of a point for a faithful primitive group action of <math>G</math>), and <math>A</math> is an [[fact about::abelian normal subgroup;3| ]][[abelian normal subgroup]] of <math>M</math> that is a [[fact about::contranormal subgroup;3| ]][[contranormal subgroup]] of <math>G</math>: <math>G</math> is generated by the conjugates of <math>A</math> in it. Then, the [[fact about::derived subgroup;3| ]][[derived subgroup]] <math>[G,G]</math> is the [[monolith]] of <math>G</math>, i.e., every nontrivial [[normal subgroup]] of <math>G</math> contains the commutator subgroup of <math>G</math>.
 ==Facts used==
@@ Line 17: / Line 17: @@
 # <math>MN = G</math>: Since <math>M</math> is core-free and <math>N</math> is nontrivial normal, <math>M</math> does not contain <math>N</math>. Since <math>M</math> is maximal, <math>MN = G</math>.
-# <math>AN</math> is normal in <math>G</math>: Clearly, <math>N \le AN \le N_G(AN)</math>. Also, <math>A</math> is normal in <math>M</math> and <math>N</math> is normal in <mah>G</math>, so <math>M \le N_G(AN)</math>. Thus, <math>MN \le N_G(AN)</math>, so <math>N_G(AN) = G</math> by step (1).
+# <math>AN</math> is normal in <math>G</math>: Clearly, <math>N \le AN \le N_G(AN)</math>. Also, <math>A</math> is normal in <math>M</math> and <math>N</math> is normal in <math>G</math>, so <math>M \le N_G(AN)</math>. Thus, <math>MN \le N_G(AN)</math>, so <math>N_G(AN) = G</math> by step (1).
 # <math>AN = G</math>: By assumption, <math>A</math> is contranormal. Thus, the only normal subgroup of <math>G</math> containing <math>A</math> is <math>G</math>. So, by step (2), <math>AN</math> is normal in <math>G</math>.
 # <math>G/N \cong A/(A \cap N)</math> is abelian: This follows from facts (1) and (2).