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If a group $G$ with order $5$ acts on a set $X$ of size $10$ what are the possible orbit partitions?

The only progress I have so far is to say that using the orbit stabiliser theorem for any $x \in X$ we have $\text{Orb}(x)$ divides $|G|=5$ so $\text{Orb}(x)=1 ~\text{or}~5$. If it is $1$ then each partition consists of a single element and the action is trivial.

If it is $5$ then the $G$ orbits partition $X$ into two sets of size $5$.

My questions are this:

Is this correct and can I do anything more?

And

The question where I got this from was an old exam paper and it was one part of a question comprised of many parts. It didn't start discussing the orbit stabiliser theorem until later parts so is there a better/more direct approach to this?

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  • $\begingroup$ For a group action two different orbits can have different sizes. $\endgroup$ – P Vanchinathan Mar 11 '17 at 15:21
  • $\begingroup$ As you rightly say the parts of the partition into orbits have size $1$ or $5$. I can make $10$ in three ways using blocks of size $1$ and $5$. $\endgroup$ – ancientmathematician Mar 11 '17 at 15:41
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Every orbit is not necessarily of the same size.

Since for $x\in X$, $|\mathcal{O} (x)|=1 $ or $5$

The possible ways to partition $X$ equal the number of ways to partition the number $10$ into sum of $1's$ or $5's$

So there are $3$ ways:
$1,1,1,1,1,1,1,1,1,1$
$5,1,1,1,1,1$
$5,5$

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