Finding elements of a cyclic set $S_{10}$ and and their orders

Question

My question read:

Show that $S_{10}$ contains elements of orders $10,20,30$. Does it contain an element of order $40$?

I am not too sure what the question is asking. Would I have to explicitly write out all the permutations in $S_{10}$ first and then find the orders for all of them?

Update: I understand I need to only show a few examples of disjoint cycles, but I am not sure how to show if order 40 is possible.

Answer 1

The answer to the exact question you ask:

Would I have to explicitly write out all the permutations in $S_{10}$ first and then find the orders for all of them?

is "no".

Hint to get you started: What's the order of $(12)(34567)$?

Answer 2

Can it be a cycle? The only cycles in $S_n$ are those of length at most $n$ hence maximum order possible $n$. So not possible to find a cyclic permutation of order 40 in S_{10}$.

Next try a permutation that is a product of two cycles, say of length $a$ and $b$. Tehn $a+b=10$; among such pairs $(a,b)$ can you find one pair with lcm$(a,b)=40$.

Next try permutations that are product of three cycles of lengths $a,b,c$. This leads to solving for $a+b+c=10$ with lcm$(a,b,c)=40$.

And so on. Now you should be able to complete the arguments.