$|G|=11$ 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!

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.
