IMC 1998 · Day 1 · P2

Let $S_n$ be the group of permutations of $\{1, 2, \dots, n\}$.

1) When $n = 3$ the proposition is obvious: if $x = (12)$ we choose $y = (123)$; if $x = (123)$ we choose $y = (12)$.

2) $n = 4$. Let $x = (12)(34)$. Assume that there exists $y \in S_n$, such that $S_4 = \langle x, y \rangle$. Denote by $K$ the invariant subgroup $$K = \{ id,\ (12)(34), (13)(24), (14)(23) \}.$$ By the fact that $x$ and $y$ generate the whole group $S_4$, it follows that the factor group $S_4 / K$ contains only powers of $\bar{y} = yK$, i.e., $S_4 / K$ is cyclic. It is easy to see that this factor-group is not comutative (something more this group is not isomorphic to $S_3$).

3) $n = 5$

a) If $x = (12)$, then for $y$ we can take $y = (12345)$.

b) If $x = (123)$, we set $y = (124)(35)$. Then $y^3 x y^3 = (125)$ and $y^4 = (124)$. Therefore $(123), (124), (125) \in \langle x, y \rangle$ – the subgroup generated by $x$ and $y$. From the fact that $(123), (124), (125)$ generate the alternating subgroup $A_5$, it follows that $A_5 \subset \langle x, y \rangle$. Moreover $y$ is an odd permutation, hence $\langle x, y \rangle = S_5$.

c) If $x = (123)(45)$, then as in b) we see that for $y$ we can take the element $(124)$.

d) If $x = (1234)$, we set $y = (12345)$. Then $(yx)^3 = (24) \in \langle x, y \rangle$, $x^2 (24) = (13) \in \langle x, y \rangle$ and $y^2 = (13524) \in \langle x, y \rangle$. By the fact $(13) \in \langle x, y \rangle$ and $(13524) \in \langle x, y \rangle$, it follows that $\langle x, y \rangle = S_5$.

e) If $x = (12)(34)$, then for $y$ we can take $y = (1354)$. Then $y^2 x = (125)$, $y^3 x = (124)(53)$ and by c) $S_5 = \langle x, y \rangle$.

f) If $x = (12345)$, then it is clear that for $y$ we can take the element $y = (12)$.