Suppose G has order 35 and |X|=17

Question

Suppose G has order 35 and |X|=17. Suppose that no point a in X is fixed by all g in G. Find the number of orbits and the size of each orbit.

I am not really sure how to go about this, the example I have been given in lectures is not of this form at all.

What's this example? Maybe there is some analogy that you can't notice at first glance:) What do you know about sizes of orbits? — savick01, Dec 02 '13 at 18:01
Emily, what do you know about sizes of orbits of a $G$-action, when $G$ has $n$ elements? — savick01, Dec 02 '13 at 18:07
I meant exactly what mixedmath wrote: that the size of an orbit is a divisor of the size of the group. — savick01, Dec 02 '13 at 20:22

score 1 · Accepted Answer · answered Dec 02 '13 at 18:11

This comes down to using the Orbit-Stabilizer Theorem, which says in short that $|Gx||G_x| = |G|$ for all $x$.

Orbit-Stabilizer tells us that the size of each orbit $Gx$ is a divisor of the size of the group $G$, so that in this case each orbit has size $1, 5, 7,$ or $35$. We clearly can't have $35$, since there aren't $35$ elements in the set. Similarly, we can't have $1$, because that would mean that every element of $g$ fixes that element.

So all the orbits are of size $5$ or $7$. But every element is in some orbit (namely, its own orbit). So we have a diophantine equation $5a + 7b = 17$ for positive $a,b$. There is exactly one solution, which is $a = 2, b = 1$.

Thus there are $3$ orbits, $2$ of size $5$ and $1$ of size $7$.

score 0 · Answer 2 · answered Dec 02 '13 at 18:07

0

Hint: Take a look at the class equation (or orbit stabilizer theorem). Use partitions of 17 and the fact that the size of the orbits divides the order of the group.

answered Dec 02 '13 at 18:07

LASV

3,108

http://www.millersville.edu/~bikenaga/abstract-algebra-2/class-equation/class-equation.html – LASV Dec 02 '13 at 18:08

Suppose G has order 35 and |X|=17

2 Answers2