Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
Same question with different numbers:… – Julian Kuelshammer Jan 30 '13 at 23:09
up vote 3 down vote accepted

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.

share|cite|improve this answer

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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.