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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

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    Doesn’t the two-element group act primitively and faithfully on itself?2012-04-03
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    Completely 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

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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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    I don't see why $|H|$ has to be odd just because $|G|$ is twice an odd number. Being a maximal subgroup doesn't mean it has the largest possible order - it just means that it's not contained in any other proper subgroup (e.g., ${\mathbb Z}/6{\mathbb Z}$ has a maximal subgroup of index 3).2012-04-04
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    @GregMartin: If H has both order and index divisible by 2, then G has order divisible by 4.2012-04-04
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    Steve, There are two trivial cases that produce the same counterexample (which is already mentioned in the comments to the question). When n=1, H is normal, and since it is core-free, H=1 and k=1. When k=1, H=1 is maximal, so n=1, and H=1 is both normal and core-free. In other words, the final contradiction is only a contradiction if n and k are bigger than 1, otherwise it produces the trivial counterexample G=2 acting regularly.2012-04-04
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    Yes, of course. I don't know why I was ignoring the $G=2$ case.2012-04-04