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A group $G$ of order $35$ act on a set $S$ that has $16$ elements. Must the action have a fixed point? Why?

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    Would you terribly mind, changing your user name to something less obnoxious? Regards,2012-04-25
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    I think your user name is fine, and you should not feel pressure at all to change it. And the action being transitive has nothing to do with it. The group can act trivially, for example, and have 16 fixed points. As the question stands now, it is meaningless.2012-04-25
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    Ah, perhaps I was too hasty. Do you mean every action **must** have a fixed point? In that case the answer is "yes", and is a more interesting question.2012-04-25
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    @KannappanSampath: No, a fixed point means an orbit of size 1.2012-04-25
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    @SteveD Right, sorry about iff. I wanted to say that it is necessary for it to be not transitive, but by no means sufficient.2012-04-25
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    How to show that the action must have a fixed point?@Steve D2012-04-25
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    @Steve D: What's the fastest way to show that any way of writing $16$ as a sum of $1$s, $5$s, $7$s, and $35$s must use at least one $1$? Of course, there are very few cases to check in this example, but what's best in general? (I hope I am not giving away too much of the answer!)2012-04-25
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    @Sexyfunction Follow up question: Can you produce a transitive action of $S_5$ on a set with $7$ elements?2012-04-25
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    @JasonDeVito Maybe something like this: if there are no 1's, then there are 5's and 7's only, but $16\equiv 1\mod 5$ while $7\equiv 2\mod 5$, so there has to be at least three 7's which is just too many.2012-04-25
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    @JasonDeVito: I just started with 2 sevens, and worked back to 0. :)2012-04-25
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    @Sexyfunction: Probably you should rewrite the question to say '_Must_ the action have a fixed point?', since that seems to be what you're actually asking.2012-04-25

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