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Exercise :

(A.1) Show that there exists an element of order $6$ in the group of permutations $S_5$.

(A.2) Show that there does not exist an element of order $8$ in the group of permutations $S_5$.

(A.3) Find all the possible orders of the elements of the group $S_5$.

Attempt :

We know that the group $S_5$ is the group of permutations of $5$ objects.

If I started in a reverse way and found every possible disjoint cycle product for $S_5$ I could answer all the $3$ questions. Take the permutations of $(12345)$. Then, starting off and writing them as a product of disjoint cycles to figure out every possible order / combination :

$$(1,2) \space \text{ of order } \space 2$$ $$(1,2,3) \space \text{ of order } \space 3$$ $$(1,2,3,4) \space \text{ of order } \space 4$$ $$(1,2)(3,4) \space \text{ of order } \space 2$$ $$(1,2,3,4,5) \space \text{ of order } \space 5$$ $$(1,2,3)(4,5) \space \text{of order } \space 6$$ $$\text{The identity permutation of order 1}$$

These are all the possible combinations of products of disjoint cycles for elements in $S_5$, from which we can see that there exists an element of order $6$ in $S_5$ and that there does not exist an element of order $8$. Also, this way (A.3) is instantly solved.

QUESTION : I guess though, that the exam exercise does not expect from a student to be solved that way. What would be another approach into avoiding the complete elaboration of disjoint cycles until (A.3) ? How would one show in another way the first two questions ?

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    $\begingroup$ I think your solution is excellent. I also don't see any other sensible way to discuss the order of permutations than to break them into disjoint cycles. In other words, it is likely that every other proof will be a (potentially sloppier) variant of yours. $\endgroup$ – user491874 Jan 21 '18 at 21:36
  • $\begingroup$ @user8734617 Thanks a lot for your comment ! You think so ? It's just weird that the exercise in built in such a way that the complete elaboration of the disjoint cycles comes last, this is why I was wondering if it expected something else from the solver ! $\endgroup$ – Rebellos Jan 21 '18 at 21:39
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    $\begingroup$ Don't forget that you also have an element of order $1$, namely the identity permutation. $\endgroup$ – Andreas Blass Jan 21 '18 at 23:49
  • $\begingroup$ @AndreasBlass Thanks for the point out, which is actually important ! I added it on my paper solution after I posted the question! $\endgroup$ – Rebellos Jan 21 '18 at 23:54
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If you play around with cycle decomposition you can show A2:

Say you have an element $x$ of order $8$. If you write the element as product of distinct cycles say $x=c_1c_2c_3c_4c_5$ (some being trivial cycles) you can show that $ord(x)=lcm(l_i)$ with $l_i$ being the length of each cycle. So for $x$ to have order $8$ you must have some cycles with order $8$ but this is non-sense since each cycle cannot be of order bigger than 5.

You can use the same reasoning for A3 too.

Comment: $S_n$ isn't the group of symmetries of the $n$-gon, $D_n$ is (except of course if $n=3$ where they are the same group).

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  • $\begingroup$ When I say that you can solve A3 using this method, I mean try counting in how many way you can partion an elemnt into disjoint cycles and for each different one find the their lcm. $\endgroup$ – Nick A. Jan 21 '18 at 21:56

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