Set of two cycles generate $S_n$

Question

How do I show that $S_n$ is generated by the set $\{(1\space i), i=2,3,...,n\}$?

Is it correct to say that since each product of elements of $\{(1\space i), i=2,3,...,n\}$ is an element of $S_n$, we know that $\{(1\space i), i=2,3,...,n\}\subset S_n$, and how can I prove that $S_n\subset\{(1\space i), i=2,3,...,n\}$?

@user149418 The set of generators in the alleged dupe is different. — AlexR, Mar 04 '15 at 20:35

AlexR · Accepted Answer · 2015-03-04T20:48:09.217

1

The inclusion $\subset$ is correctly justified. The other way ($\supset$) can be proved by giving an explicit construction of $$S_n \ni \pi = (1\ i_1)(1\ i_2)\ldots(1\ i_k)$$ Where $k$ is finite, $\pi$ is a given permutation and the $i_j$ can be chosen. For this, look at products $(1\ i_k)\ldots(1\ i_1)(1\ i_k)$. What are they?

For example $(1\ 2)(1\ 3)(1\ 2) = \;?$

Observe the following:

$$(1\ i_k)\ldots(1\ i_1)(1\ i_k) = (i_1\ i_2\ \ldots\ i_k)$$
So we can generate any cycle with this construction. Since every permutation is a product of disjoint cycles, we can decompose a permutation into its cycles and apply the construction on each cycle. The product of these is a representation of $\pi$ as a product of the swaps $(1\ i)$ for some $i$.
For example $\pi = (1 3 2)(4 5) = (12)(13)\cdot (1 5)(1 4)(1 5) = (12)(13)(15)(14)(15)$

edited Mar 04 '15 at 20:48

answered Mar 04 '15 at 20:37

AlexR

25,110

are they a permutation $(1,i_1,i_2,...,i_k)$? – Stan_Allen Mar 04 '15 at 20:39
@Stan_Allen Almost. They form a cycle, but wich cycle? – AlexR Mar 04 '15 at 20:39
$(12)(13)(12)=(23)$ – Stan_Allen Mar 04 '15 at 20:40
@Stan_Allen Correct. Do you notice what happens for the general case? Do you see how we can now write an arbitrary permutation? – AlexR Mar 04 '15 at 20:41
not exactly.... – Stan_Allen Mar 04 '15 at 20:41
Okay see what $(1\ 2)(1\ 3)(1\ 4)(1\ 2)$ gives you. Note that every permutation can be written as the product of disjoint cycles. – AlexR Mar 04 '15 at 20:42
$(12)(13)(14)(12)=(243)$ – Stan_Allen Mar 04 '15 at 20:43
does this mean that the answer to your general example is $(i_1, i_k,i_{k-1},...,i_2)$? – Stan_Allen Mar 04 '15 at 20:46
@Stan_Allen Yes! I decided to re-index so $$(i_1\ \ldots\ i_k) = (1\ i_k)(1\ i_{k-1})\ldots(1\ i_1)(1\ i_k)$$ – AlexR Mar 04 '15 at 20:49
I get it now! thanks a lot for all the help! :) – Stan_Allen Mar 04 '15 at 20:51

Set of two cycles generate $S_n$

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