Assume we know that $S_n=⟨(12),(13),...,(1n)⟩$, then prove $S_n=⟨(12),(23),...,((n-1)n)⟩$
I didn't understand this proof below:
For $i=2$ at $(1 \ i)$ so $(1 \ i) \in S_n$, then assuming that $(1 \ i)$ for all $i\in\{1,...,n\}$ we have
$(1 \ i+1)=(1 \ i)(i \ i+1)(1 \ i)$ then $(1 \ i+1) \in S_n$ so $(1 \ i) \in S_n$ really hold.
What I didn't get:
- Why does this factorization imply to $(1 \ i+1)\in S_n$?
- If the reason for that is $(1 \ i) \in S_n, (i \ i+1) \in S_n$ implies to the "product" belonging to $S_n$, then why would I not just write $(1 \ i+1)=(1 \ i)(i \ i+1)$?
$⟨(12),(23),...,((n-1)n)⟩$ means all elements which are generated by the product of some terms of $(12),(23),...,$ and $((n-1)n)$.
Now $(13)=(12)(23)(12)$ so $(13)$ is an element of $⟨(12),(23),...,((n-1)n)⟩$ and we can consider $(13)$ as one of the generators.
Now $(14)=(13)(34)(13)$ so $(14)$ is an element of $⟨(12),(23),...,((n-1)n)⟩$ and we can consider $(14)$ as one of the generators.
We can continue this by induction, and, with the knowledge of $(1\ i) \in⟨(12),(23),...,((n-1)n)⟩$, we realize $(1\ i+1)=(1\ i)(i\ i+1)(1\ i) \in ⟨(12),(23),...,((n-1)n)⟩$
Finally, since $(1\ i) \in ⟨(12),(23),...,((n-1)n)⟩$ for all $i$ and $S_n=⟨(12),(13),...,(1n)⟩$, then $S_n=⟨(12),(23),...,((n-1)n)⟩$.