# Smallest Subgroup of Sn

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• Dec 9th 2009, 11:43 AM
RoboMyster5
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 http://www.mathhelpforum.com/math-he...94de62bf-1.gif such that http://www.mathhelpforum.com/math-he...883c0ba4-1.gif,
Above I proved that (12)(1...n) generates Sn so therefore
http://www.mathhelpforum.com/math-he...33c6fce1-1.gif
How do I prove that
http://www.mathhelpforum.com/math-he...6fc81d6c-1.gif. so I can establish that G=Sn??

AHHHH

• Dec 9th 2009, 01:13 PM
aliceinwonderland
Quote:

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 http://www.mathhelpforum.com/math-he...94de62bf-1.gif such that http://www.mathhelpforum.com/math-he...883c0ba4-1.gif,
Above I proved that (12)(1...n) generates Sn so therefore http://www.mathhelpforum.com/math-he...33c6fce1-1.gif
How do I prove that http://www.mathhelpforum.com/math-he...6fc81d6c-1.gif. 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.
• Dec 9th 2009, 01:55 PM
RoboMyster5
Wow I think you just made this make complete sense...I applaud you. Thank you.