I want to ask for a hint to this problem: G acts primitively, faithfully on A. |A| is even, show that |G| is divisible by 4
G acts primitively, faithfully on A. |A| is even, show |G| is divisible by 4
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group-theory
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0Completely ignorant of group theory, but I'm well aware that $2^2=4$. Does that have anything to do with the issue at hand? – 2012-04-03
1 Answers
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Let's translate this to a question about subgroups of $G$. The stabilizer of this transitive action is a subgroup, let's call it $H\le G$. To say the action is primitive is to say $H$ is a maximal subgroup, to say it is faithful is to say there are no normal subgroups contained in $H$, and by the orbit-stabilizer theorem, $[G:H]=|A|$. Now $|A|$ is even, so $[G:H]=2n$. If $|G|$ is not divisible by $4$, then $n$ is odd and $|H|$ is odd. Letting $k=|H|$, we have $|G|=2nk$, which is twice an odd number. It is easy to then see (using the regular representation) that $G$ has a subgroup of index $2$, call it $N$ [note that $N$ is normal]. We also have $H\le N$, contradicting $H$ being maximal and non-normal.
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0Yes, of course. I don't know why I was ignoring the $G=2$ case. – 2012-04-04