0
$\begingroup$

Prove by induction $\langle (1,2),(2,3),...,(n-1,n) \rangle = S_n$ for $n\geq 2$.

Base case: $n=2$ then $\langle (1,2) \rangle =\{(),(1,2)\} =S_2$.

Assume that: $\langle (1,2),(2,3),...,(k-1,k) \rangle = S_k$ for $k>2$ we want to show this implies $\langle (1,2),(2,3),...,(k-1,k), (k,k+1) \rangle = S_{k+1}$.

I have no idea how to prove this implication though. Any help?

2 Answers 2

0

Hint: Because every permutation is a product of transpositions , it is enough to prove that $\langle (1,2),(2,3),\ldots,(n-1,n) \rangle $ contains all transpositions $(ij)$.

To complete the induction, show how to write $(k,n+1)$ in terms of $S_n$ and $(n,n+1)$. You'll find that a simple conjugation works.

  • 0
    Okay so let $(ij)$ be a given transposition in the case $i=j$ this is the identity and in the case $j=i+1$ this is clearly contained. Now if $(i,j)$ is neither of those cases and $i,j \in \{1,2,3,...,k\}$ an suppose without loss that $i$(ij)=(i,i+1)(i+1,i+2),...,(j-1,j)$ which is also clearly contained. I see this but how is this enough to conclude what I want. (I kind of see it intuitively because we can just take the identity map and shift it around enough to get anything we want right?) how can I write this formally? – 2017-01-27
0

We know that $\sigma = (12), \;\tau = (123 \cdots n)$ can form the group $S_n$ by transposition.

$(\sigma)^k \circ \tau \circ (\sigma^{-1})^k = (k+1, k+2)$ must be a $n$ cycle.