1
$\begingroup$

There is a problem in Problems in Group Theory by J.D.Dixon :

2.51 If a permutation group $G$ contains a minimal normal subgroup $N$ which is both transitive and abelian, then $G$ is primitive.

He answered it amazingly:

Let $G_\alpha $ be a stabilizer of $G$. It is sufficient to show that $G_\alpha $ is a maximal subgroup of $G$. Let us suppose that, on the contrary, $G_\alpha $ is not a maximal subgroup of $G$. Then, there is a proper subgroup $K$ of $G$ properly containing $G_\alpha $. Since $K=K∩G_\alpha N=G_\alpha$ some $x≠1$ in $N$ lies in $K$. Therefore $K∩N = M$ is a nontrivial...

Is it possible we have an errata in $K=K∩G_\alpha N=G_\alpha$ cause of wrong typing or printing? Clearly, there would be an inconsistence with his assumption, and if so, which ones of the following would be the correct one:

$K⊇K∩G_\alpha N⊇G_\alpha$, or $K=K∩G_\alpha N⊇G_\alpha$

I will be so pleased if someone points the right one. I couldn’t find an erratum for this great book.

1 Answers 1

5

It is well known that in this case $G=G_\alpha N$, so, clearly, $K=K\cap G=K\cap G_\alpha N$. The second part of the equality is clearly wrong, or we would have had a contradiction alredy. What should be there (and what is really used to say that there exist a $x\neq 1$ such that $x\in N\cap K$) is $K\cap G_\alpha N\supsetneqq G_\alpha$. (or, simply, $K\cap G_\alpha N\neq G_\alpha$ since we already know that $G_\alpha\subset K$)

  • 0
    Thanks so much dear Dennis.2012-06-19