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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.

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    Do you know how to write permutations in cycle notation? Do you know how to find the order of a product of disjoint cycles? https://en.wikipedia.org/wiki/Permutation#Cycle_notation – 2017-02-11
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    @EthanBolker yes to both questions – 2017-02-11

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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 $(12)(34567)$?

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    it would be the lcm of 2, 5 which is 10 – 2017-02-11
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    So now you have part of the answer. You found an element of order 10. You should be able to finish by yourself. – 2017-02-11
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    oh okay so this is all I have to do? I have to just pick the numbers in S_10 to create these cycles and get the orders I want? – 2017-02-11
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    That's all you have to do. – 2017-02-11
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    Oh but then say for element of order 10, I have to find all the possible combinations? Wouldn't the above example be wrong because 8 9 and 10 in S_10 are not included? – 2017-02-11
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    No. $(12)(34567) = (12)(34567)(8)(9)(X)$ is an element of $S_{10}$. It just happens to fix $8$, $9$ and $X$. So go ahead and finish. Order $40$ may require some argument. Good night. – 2017-02-11
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    Okay, thank you. Then I will only find a couple examples for each order. – 2017-02-11
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    I am sorry I worked on the problem and I showed a few examples for the orders 10 20 and 30. I see know after trying that order 40 is not possible but I am not too sure how to say this in words. Would just showing an example that it fails work? – 2017-02-13
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    No. An example is enough to show existence but a failed example doesn't show nonexistence. Look at the divisors of $40$. Show that you can't get the lcm of the lengths of disjoint cycles to reach $40$ when you have just ten elements to play with. – 2017-02-13
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    Yes, I was thinking of saying to get lcm equal 40 you would need lcm ( 5,8) but this does not work because there are only 10 elements – 2017-02-14
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    @Sam So that's the answer. – 2017-02-14
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    This would be enough to say why this does not work. Wouldn't I need to state for example the prime factorization of 40? – 2017-02-15
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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.

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    From what you said for your first statement then wouldn't that go again finding an element of order 20? – 2017-02-11