Some problem about symmetric group

Question

Given a symmetric group $S_n$, it is easy to see that $S_n$ can be generated by two permutataions: $(12)$ and $(123...n)$, since we can rewrite any transposition $(k\ k+1) = (123...n)^k(12)(123...n)^{-k}$. And since the $S_n$ can be generated by the transpositions $(12),(23),..,(n-1\ n)$, we conclude that the claim is true.
What I am curious about is that whether any n-cycle with the transposition $(12)$ can generate the whole symmetric group $S_n$.
I get this problem on my abstract algebra class recently, which my teacher proposed it but could not give a right answer in time.

Answer 1

Here is a more elementary version of this answer.

Consider $S_4$ permuting the sequence $a,b,c,d$ and the cycles $(1234)$ and $(13)$. If we start with $a,b,c,d$, the letters $a$ and $c$ will always be $2$ spaces apart from each other. Clearly $(1234)$ cannot change how far apart they are, and permuting the first and third element will either a) not affect where $a$ and $c$ are or b) switch them with each other, which doesn't change the distance between them. This is because it is impossible for only one of $a$ and $c$ to be in the positions $1$ or $3$.

This might look like a circular argument, but we have shown that if, after move $n$ the letters $a$ and $c$ are not $2$ apart from each other, then must not have been 2 apart from each other before move $n$ either. So by induction they must always be $2$ apart. Hence the group generated by $(1234)$ and $(13)$ e.g. has no element that can map $a,b,c,d$ to $a,b,d,c$, so it cannot be the full symmetric group.

Answer 2

Here is a counterexample. The subgroup of $S_4$, given by $$ H=\langle (12),(2413)\rangle $$ has order $8$, hence cannot be isomorphic to $S_4$. This has been proved here, with $2$ and $3$ interchanged. In fact, $H\cong D_4$, the dihedral group.

In general, if $n=p\ge 3$ is prime, then $S_p$ is generated by any $2$-cycle and any $n$-cycle.

Reference:

Let $a$ be a $p$-cycle in $S_p$, and let $b$ be a transposition in $S_p$. Show $S_p$ is generated by $a$ and $b$.

Answer 3

We can prove a general result:

$(ab)$ and $(a_1a_2\cdots a_n)=:\sigma$ generates $S_n$ iff the distance between $a$ and $b$ in $(a_1a_2\cdots a_n)$ is coprime to $n$.

Note that the distance between $a$ and $b$ can be interpreted in two different ways: the one from $a$ to $b$ and the one from $b$ to $a$, but since their sum is $n$, so one of them is coprime to $n$ iff both of them are coprime to $n$, hence we can use either in the above proposition.

Without loss of generality, let's assume $a_i=i, a=a_1$, then the distance between $a$ and $b$ is just $b-1$. We divide $\{1, \cdots, n\}$ into mod $d$ classes, where $d>1$ is a nontrivial common divisor of $n$ and $b-1$. We shall show both $(1b)$ and $\sigma$ leave the mod $d$ classes invariant, hence they cannot generate the full $S_n$.

Note that $1$ and $b$ are in the same class as $d\mid b-1$, so $(1b)$ will leave each class invariant. As for the action of the $n$-cycle $\sigma$, we have $\sigma(x)\mod n\equiv (x+1)\mod n$, so $\sigma(x)\mod d\equiv (x+1)\mod d$ since $d\mid n$. Therefore if $x\equiv y \mod d$ then $\sigma(x)\equiv \sigma(y)\mod d$.

If $\gcd(n, b-1)=1$, consider $\sigma^{b-1}$ which sends $1$ to $b$, and it satisfies

$$\sigma^{b-1}(x)\mod n \equiv x + b - 1\mod n$$

In the cycle of decomposition of $\sigma^{b-1}$, if the cycle that contains $1$ has length $m$, then $m(b-1)\equiv 0 \mod n \Rightarrow m = n$. So $\sigma^{b-1}$ is still a $n$-cycle that starts with $(1, b, \cdots)$, therefore it generates $S_n$ since $(12)$ and $(123\cdots n)$ generate $S_n$.