# $|G|=11$ operates on $\mathbb{Z} / 5\mathbb{Z}\times \mathbb{Z} / 5\mathbb{Z}$ implies at least one point fixpoint

Can someone please help me with the following: A group of order $11$ operates on $\mathbb{Z} / 5\mathbb{Z}\times \mathbb{Z} / 5\mathbb{Z}$. I have to show that it has at least one fix point. Can someone help me please with it or give me at least some hint ?

I have an exam tomorrow so please give me some hint!

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Same question with different numbers: math.stackexchange.com/questions/136557/… – Julian Kuelshammer Jan 30 '13 at 23:09

## 2 Answers

When a finite group acts on an object, the object is partitioned into orbits, where each orbit has cardinality that divides the order of the group. In this case the only possible divisors of $11$ are $11$ and $1$. Now ${\mathbb Z}/5{\mathbb Z} \times {\mathbb Z}/5{\mathbb Z}$ has cardinality $25$ which is not a multiple of $11$, so you can't just have $11$'s; you have either two orbits of size $11$ and three fixed points, or one orbit of size $11$ and $14$ fixed points, or $25$ fixed points.

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Robert's answer already wrapped up this question, but you may be interested in proving the following general proposition, which solves at once your question:

Proposition: If $\,G\,$ is a finite $\,p-$group and $\,X\,$ is a non-empty finite set upon which $\,G\,$ acts, then

$$|X|=\left|X^G\right|\pmod p\;\;,\;\;\text{with}\;\;X^G:=\{x\in X\;gx=x\,\,\forall g\in G\}=\{x\in X\;;\;|\mathcal Orb(x)|=1\}$$

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