Can a permutation of 9 elements with order 4 have no fixed points?

Question

My thinking is

With 9 elements, having order 4 means either a single orbit of length 4 and zero, one, or two orbits of length 2; or two orbits of length 4.
in both cases, the number of fixed points (not in one of the orbits) is odd, so can't be 0

Am I missing something?

(I am an amateur here - my math is from engineering, so I know some terms but not the whole theory)

I thought so, but is that rigorous enough to say "I proved that it must have a fixed point"? — laugh salutes Monica C, Nov 04 '17 at 12:35
It was a "show a permutation of order 4" question in my daughter's homework. Turned out to be a mistake in the question; it was updated to "order 12" after a while. Which became a simple task (2,3,4). — laugh salutes Monica C, Nov 14 '17 at 17:50

score 0 · Accepted Answer · answered Nov 04 '17 at 12:34

0

You are perfectly right!

Another way to think of it is to look at the cycle decomposition of a permutation of $9$ elements (see e.g. here).

Having a permutation of order 4 is equivalent of saying that its disjoint cycles can only have length $1,2$ or $4$. If you exclude fix points (i.e. cycles of length $1$), there is no way to get to $9$ elements only using cycles of length $2$ or $4$.

answered Nov 04 '17 at 12:34

chickenNinja123

651

Thanks for the link. It helped me make the claim with some confidence. – laugh salutes Monica C Nov 14 '17 at 17:43

Can a permutation of 9 elements with order 4 have no fixed points?

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