Using a simple application of Sylow's theorems, you want to show that there is only one Sylow 11-subgroup. Why is this sufficient?

For (a) I ask must ask you a question...what does an element of order 3 in look like (in cycle notation)? Can you count how many elements have this form?

For (b), you know that there are ` ' elements of order 3. Each of these must appear in some Sylow 3-subgroup of order 3, and so they must appear in pairs in this subgroup. So your answer is (do you understand why this is)?

You should check that this answer is valid by plugging in Sylow's Theorems (they do not give you the actual answer in this case, just a couple of possible answers).

For (c), what is the order of a Sylow 2-subgroup? I shall call this number ` '. Now, can you think of a well-known result about groups of order `m'? For the next bit, every group of order is a Sylow 2-subgroup (why?), and as all such groups are conjugate they are all isomorphic. Therefore, to finish the question you need to find a subgroup (any subgroup!) of of order ` '. What does this group look like?

Hmm...(a) and (b) are actually false, but I suspect you are only looking at finite groups and so are disproved by simple order arguments. (For (b), a counter-example is the integers...I can't think of a counter-example for (a) off-hand EDIT: Take the infinte cross-product of with itself. This is an infinitely generated abelian group. Then, quotient out one of the 's...).

As a hint for (c), `Cayley's Theorem'.

For (d), I believe your first counter-example comes at order 6. However, you should note that for . So...

And I will leave you to solve (e). You proof should start `Let ...' (again, it is false).