What Is the Largest Order of an Element in S_{10}?

[ElDavidas]

I was in a tutorial today and was asked

"What is the largest order that an element of $$S_{10}$$ can have?"

I thought the answer was 10! but I've been told this is wrong. Can someone help me out with what's going on? I thought you calulated the order by the formula:

$$|S_n| = n!$$

 

[Muzza]

You were asked about the order of an element, not the order of the group.

There can't be an element of order 10! in S_10, because then S_10 would be abelian (even cyclic).

Do you know that any permutation can be written as the product of disjoint cycles?

 

[JasonRox]

I noticed that a lot of people can't answer these questions when asked.

The question that Muzza just asked is something you should know to answer the question you want to know.

 

[matt grime]

Ahem, this seems that it is also a matter of English and presumption.

For the following question:

Let G be a group of order m, what is the largest order an element can have?

Then the correct answer really is m, since all elements have order dividing m and there is always a cyclic group of order m.

However, just because something can happen doesn't mean it does happen. If we're given the extra information that G is actually S_n and n!=m, then, we can get a *better* answer, and indeed we can explicitly say what all permissible orders of elements are.

Can is a bad word, in this question, or many questions. The better phrase would be: what is the largest order of an element of S_n.