$(12) \notin \langle (132), (123456) \rangle$ How can I prove/disprove that as an element of $S_6, $ $(12) \notin \langle (132), (123456) \rangle$. We can use GAP to check this, but by hand it looks not so obvious. 
Let $r=(123456)$ is odd and $s=(132)$ is even , so any word in {$r^{\pm},{s^{\pm}}$}, can give an odd permutation as well as even. 
 A: GAP tells me this:
gap> G := Group((1,3,2), (1,2,3,4,5,6));
Group([ (1,3,2), (1,2,3,4,5,6) ])
gap> (1,2) in G;
true
gap> GeneratorsOfGroup(G);
[ (1,3,2), (1,2,3,4,5,6) ]
gap> Factorization(G, (1,2));
(x2^2*x1^-1)^2*x2

A: By hand:
$$(1\ 2)=(1\ 2\ 3\ 4\ 5\ 6)^4(1\ 3\ 2)(1\ 2\ 3\ 4\ 5\ 6)^4(1\ 3\ 2)(1\ 2\ 3\ 4\ 5\ 6)^5$$
This is how I worked it out:
$$(1\ 2\ 3\ 4\ 5\ 6)=(1\ 2)(2\ 3\ 4\ 5\ 6)$$
so
$$(1\ 2)=(1\ 2\ 3\ 4\ 5\ 6)(2\ 3\ 4\ 5\ 6)^{-1}=(1\ 2\ 3\ 4\ 5\ 6)[(2\ 3\ 4)(4\ 5\ 6)]^{-1}$$
and of course
$$(2\ 3\ 4)=(1\ 2\ 3\ 4\ 5\ 6)(1\ 2\ 3)(1\ 2\ 3\ 4\ 5\ 6)^{-1}=(1\ 2\ 3\ 4\ 5\ 6)(1\ 3\ 2)^{-1}(1\ 2\ 3\ 4\ 5\ 6)^{-1}$$
and
$$(4\ 5\ 6)=(1\ 2\ 3\ 4\ 5\ 6)^3(1\ 2\ 3)(1\ 2\ 3\ 4\ 5\ 6)^{-3}=(1\ 2\ 3\ 4\ 5\ 6)^3(1\ 3\ 2)^{-1}(1\ 2\ 3\ 4\ 5\ 6)^{-3}$$
etc.
A: $(1\,3\,2)$ means that you can do a cyclic permutation of any three neighbors in the long cycle, by conjugating with an appropriate multiple of $(1\,2\,3\,4\,5\,6)$.
In particular you can move one element two places to the left or to the right in the cyclic order. Since there is an odd number of other elements to jump over, moving one element all the way around the large cycle in steps of two will leave it swapped with its neighbor.
Thus, we can make $({}^1_2 {\,}^2_3 {\,}^3_1 {\,}^4_4 {\,}^5_5 {\,}^6_6) = (1\,3\,2)^2$ and then (multiplying on the right with an appropriate conjugate) $({}^1_2 {\,}^2_3 {\,}^3_4 {\,}^4_5 {\,}^5_1 {\,}^6_6)$ and then $({}^1_1 {\,}^2_3 {\,}^3_4 {\,}^4_5 {\,}^5_6 {\,}^6_2)$. Finally rotate right once to get $({}^1_2 {\,}^2_1 {\,}^3_3 {\,}^4_4 {\,}^5_5 {\,}^6_6) = (1\,2)$
