Can we generate $S_n$ by all permutations with no fixed point?

Question

I'm working on the following problem and I've made some promising progress:

Show that for $n>3$, each element of the symmetric group $S_n$ is a product of permutations which have no fixed point. (ie $n$-cycles)

I've figured this out for $n$ even. Heres how you'd do it. Take $n=6$ as an example. Note that $$(1 2 3 4 5 6)^2 = (1 3 5)(2 4 6) = (1 2)(1 3 5 2 4 6)$$

So $$(1 2) = (1 2 3 4 5 6)^2(6 4 2 5 3 1)$$

Now that we've generated one 2-cycle, we can generate all of them in the same way. I'm having trouble getting this idea to work for $n$ odd though. I'm curious to see if anyone else knows how to do it either using this idea or some other idea.

Source: Fall 1996

Answer 1

The subgroup $G\subset S_n$ generated by elements with no fixed points is normal, as $g^h(x) = x$ iff $g(h x) = hx$. Write down an odd such element to prove the result for $n\geq 5$; the case $n = 4$ can be handled directly.

Answer 2

Recently I showed a stronger statement. if n > 3, then any permutation is the product of a derangement (permutation with no fixed point) and an n-cycle (which is also a derangement). Let $S_n$ permute {1,2,...,n}. First I prove that every permutation is conjugate to a non-successor permutation; i.e., a permutation p such that for no i is p(i)=i+1 (and p(n) is not 1). For n=4, every permutation is conjugate to one of (), (13), (13)(24), (132), and (1432) all of which are non-successor. If n>=5, if p is an n-cycle, select i relatively prime to n and not 1. Then (1 1+i ... 1+(n-1)i) is non-successor. If p has a fixed point, then by induction p|{1,2,...n-1} is in $S_{n-1}$ and is non-successor. Relabel (that is the same as conjugating) {1, ..., n} so that n is the fixed point. Then p|{1,...,n-1}(n) is non-successor. If p contains a transposition, relabel so the transposition is (n-1 n). p|{1,...,(n-2)} by induction is non-successor (if it is in $S_3$, use (132))so p|{1...(n-2)}(n-1 n) is non-successor. If p contains a 3 cycle but no 2-cycle, a similar argument can be made, noting that n in this case has to be >=6. If all cycles of p are 4 or higher, then break p into two chunks of size 4 or higher, express the chunks as non-successor by induction and put them back together. // From this we show that any permutation p is the product of a derangement and an n-cycle. p = x*{1 n n-1 n-2... 2}. For all i, x does not take i into i+1 since it is non-successor (conjugate if necessary). Therefore, multiplying by the n-cycle does not take i into i. Therefore p is the product of a derangement and an n-cycle.