A group $G$ of order $35$ act on a set $S$ that has $16$ elements. Must the action have a fixed point? Why?
A question about the fixed point and group action.
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group-theory
finite-groups
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5Would you terribly mind, changing your user name to something less obnoxious? Regards, – 2012-04-25
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7I 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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0Ah, 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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1@KannappanSampath: No, a fixed point means an orbit of size 1. – 2012-04-25
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0@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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0How to show that the action must have a fixed point?@Steve D – 2012-04-25
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4@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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0@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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0@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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0@JasonDeVito: I just started with 2 sevens, and worked back to 0. :) – 2012-04-25
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1@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
1 Answers
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(Essentially Jason's argument.)
Yes, every action of this group should have a fixed point.
Size of orbits divide the order of the group (comes from Orbit-Stabilizer Lemma). So, your orbits should be of size $1$, $5$, $7$ or $35$.
Now, since the set is of cardianlity $16$, we cannot have an orbit of size 35. Now, suppose there we no fixed points, then:
for $n, m \ge 1$, $7n+5m=16$ has integer solutions in $n$ and $m$.
Now one should argue this is not the case.
For $n \ge 2$, $16=7n+5m \ge 14+5m$ which would mean, $5m \le 2$ which is impossible with $m \in \Bbb{Z}$. So, $n=1$. But, then this means, $5m=11$ which is again absurd as $m \in \Bbb{Z}$.
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2You have *completely* rewrote your answer to the one already given by Jason DeVito in the comments, or am I missing something? – 2012-04-25
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0Yes, right. @Vadim. I have completely rewritten that. I realized that hint would not work. I did read Jason's comment. So, this is essentially his argument. – 2012-04-25
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0Ok, no problem. Just to clarify. – 2012-04-25
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1Will the downvoter care to explain? – 2012-04-29
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0I made a comment that indicates that I might be the downvoter, but I am not... So, must be someone else, who did not bother to explain. – 2012-05-10