6

For $n \geq 4$, I want to show that $S_n$ has a minimal set of generator of size $k$ for $k \in \{2, \cdots, n - 1 \}$, and also that it does not have a minimal set of generator of size $n$.

So I already know that $S_n$ can be generated by $\{ (1, \cdots, n), (1, 2) \}$ and also a set of transpositions, but I am not sure how to proceed from here. Can I somehow extend from $k = 2$ to more?

Trash Failure
  • 771
  • There's actually a theorem that (for any group) the set of natural numbers $k$ such that there exists a minimal generating set of cardinality $k$ has "no gaps", i.e. if there are minimal generating sets of cardinalities $a, b$ and $a < c < b$, then there exists a minimal generating set of cardinality $c$. So this solves (part of) your problem if you know the proof of the irredundant basis theorem! – diracdeltafunk Jan 21 '21 at 06:17
  • 1
    This seems to be the same question – diracdeltafunk Jan 21 '21 at 06:24
  • 1
    @diracdeltafunk It's not an exact duplicate, because it does not address the existence of minimal generating sets for all $k \in {2,\ldots,n-1}$, but that's very easy to prove directly for $S_n$: just take ${(1,2),(2,3),\ldots,(k-1,k),(k-1,k,k+1,\ldots,n)}$. – Derek Holt Jan 21 '21 at 07:58
  • @DerekHolt How do I know if it is a minimal? – Trash Failure Jan 21 '21 at 12:08
  • Sorry I got it wrong! It should be ${(1,2),(2,3),\ldots,(k-1,k),(k,k+1,\ldots,n)}$. You know that it is minimal by proving it. Note that if you leave one of the generators out, then the group generated by the rest does not act transitively on ${1,2,\ldots,n}$, so it cannot be the whole of $S_n$. – Derek Holt Jan 21 '21 at 13:11

0 Answers0