1. Smallest Subgroup of Sn

Good afternoon fellow math goers.
I entered this problem a week ago and with helpful hints have made what I think is progression on this rather challenging problem. I need some advice on my proof though so any more help would be greatly appreciated.

Problem: Prove that the smallest subgroup of Sn containing (1, 2) and (1, 2, . . . , n) is Sn.

Conjunction:
(12...n)(12)(12...n)^-1=(23)
(12...n)(23)(12...n)^1=(34)
(12...n)(34)(12...n)^-1=(45)
WLOG (12...n)(n-2 n) (12...n)^-1=(n-1 n)
(12...n)^n-1 (n-1 n)(12...n)^n-1=(n1)

So essentially this gives us a difference k=0 and that we proved its a generator of Sn correct? Now should I use induction and do k+1 terms or should I do the following:

Suppose you have some group such that ,
Above I proved that (12)(1...n) generates Sn so therefore

How do I prove that
. so I can establish that G=Sn??

AHHHH

2. Originally Posted by RoboMyster5
Good afternoon fellow math goers.
I entered this problem a week ago and with helpful hints have made what I think is progression on this rather challenging problem. I need some advice on my proof though so any more help would be greatly appreciated.

Problem: Prove that the smallest subgroup of Sn containing (1, 2) and (1, 2, . . . , n) is Sn.

Conjunction:
(12...n)(12)(12...n)^-1=(23)
(12...n)(23)(12...n)^1=(34)
(12...n)(34)(12...n)^-1=(45)
WLOG (12...n)(n-2 n) (12...n)^-1=(n-1 n)
(12...n)^n-1 (n-1 n)(12...n)^n-1=(n1)

So essentially this gives us a difference k=0 and that we proved its a generator of Sn correct? Now should I use induction and do k+1 terms or should I do the following:

Suppose you have some group such that ,
Above I proved that (12)(1...n) generates Sn so therefore
How do I prove that . so I can establish that G=Sn??

AHHHH

From my understanding, you used the theorem that S_n is generated by n-1 transpositions (1,2), (2,3), ..., (n-1, n). You showed that (1,2) and (1,2, ... , n) indeed generate those n-1 transpostions, which shows that $\displaystyle S_n \subseteq G$. We see that G (a set generated by (1,2) and (1,2,...,n) ) is just the set of permutations of {1,2, ..., n}. By the definition of the symmetric group of degree n, it forces $\displaystyle G \subseteq S_n$. Thus G=S_n, a symmetric group of degree n.

EDIT (some additional remarks): G must contain (1,2) and (1,2, ... , n). Since G is the group by hypothesis, G must contain a set generated by (1,2) and (1,2, ... ,n). Since our choice of G should be minimal, if we show that a set generated by (1,2) and (1,2, ... ,n) is the group, the proof is completed.

3. Wow I think you just made this make complete sense...I applaud you. Thank you.