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Show that $S_n$ is generated by the set $ \{ (12),(123\dots n) \} $.

I think I can see why this is true. My general plan is (1) to show that by applying various combinations of these two cycles you can get each transposition, and then (2) to show that each cycle is a product of transpositions.

I'm just having trouble on the first step. Any ideas?

Martin Sleziak
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Dan H
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    If $h=(12\ldots n)$, the $(12)^{h}=(23)$, $(12)^{h^2}=(34)$, etc. –  Oct 09 '13 at 20:16
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    I'm confused by the exponents. Can you clarify? Thanks – Dan H Oct 09 '13 at 20:33
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    @Dan: Conjugation is often written as exponentiation, so $x^y$ means $y^{-1} x y$. It does satisfy the suggested identities: $(xy)^z = x^zy^z$ and $x^{yz} = (x^y)^z$. More general group automorphisms are also often written with similar notation. –  Oct 11 '13 at 07:37
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    And it's useful to actually think in this way: e.g. conjugation by $h$ turns out to be a fairly natural operation on cycles (it adds one to each number appearing in the cycle). What might conjugation by $(23)$ do? –  Oct 11 '13 at 07:43
  • (depending on your ordering conventions, $x^y$ might mean $yxy^{-1}$ -- I don't recall which one is standard) –  Oct 11 '13 at 07:47

3 Answers3

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Let $c = (1, 2, \dotsc, n)$. We see that \begin{align*} c (1, 2) c^{-1} &= (2, 3) \\ c (2, 3) c^{-1} &= (3, 4) \\ &\vdots \\ c (n-2, n-1) c^{-1} &= (n-1, n), \end{align*} so that $(i, i+1) \in \langle (1, 2), c \rangle$ for all $1 \leq i \leq n-1$. Next, we have \begin{align*} (2, 3) (1, 2) (2, 3)^{-1} &= (1, 3) \\ (3, 4) (1, 3) (3, 4)^{-1} &= (1, 4) \\ &\vdots \\ (n-1, n) (1, n-1) (n-1, n)^{-1} &= (1, n), \end{align*} so that $(1, i) \in \langle (1, 2), c \rangle$ for all $1 \leq i \leq n$. Choose any $1 \leq i < j \leq n$, then $$ (i, j) = (1, i) (1, j) (1, i)^{-1} \in \langle (1, 2), c \rangle. $$ Therefore, $\langle (1, 2), c \rangle$ contains all transpositions. Hence, $\langle (1, 2), c \rangle = S_n$.

tylerc0816
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  • I was thinking that would be true so that I could get that the set is equivalent to all transpositions, and every cycle can be written as a product of transpositions, but I'm not sure why its true. – Dan H Oct 09 '13 at 20:31
  • I edited it to include an answer now. – tylerc0816 Oct 09 '13 at 20:42
  • Isn't $c(1, 2)c^{-1} = (1,n)$? Doesn't change anything to the solution though. – Quotenbanane Apr 29 '21 at 22:30
  • @Quotenbanane yes you can generate (1n) and then you have to look for another n order element to generate one more 2 cycles, it's pretty cumbersome, rather just find a way to generate (12) , (13)... (1n) then (23), (24),... It's easy to get track here. – M Desmond Oct 14 '21 at 08:23
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A simpler proof:

We know $S_n=\langle(12),(23),\ldots,(n-1,n)\rangle$. As tylerc0816 has pointed out $$(12),(23),\ldots,(n-1,n)\in\langle(12),(12\ldots n)\rangle.$$

The smallest group containing all adjacent traspositions is $S_n$. Therefore, $S_n=\langle(12),(12\ldots n)\rangle$.

amWhy
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moqui
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To show $S_n$ is generated by $K=\{(12),(1 2 3...n)\}$ we first show that $S_n$ is generated by $H=\{(12),(23),...,(n \ n-1)\}$
Since every permutation is product of transposition, it suffices to show that $(ij)$ such that $i<j$ belongs to $H$
Let $$(ij)=(i\ i+k) \ \forall \ k \geq 1$$
Proof by induction
for $k=1$ $(i \ i+1) \in H$
assume it true for $k$
Show for $k+1$
$$(i \ i+k+1)=(i \ i+1)(i+1 \ i+1+k)(i \ i+1)$$
clearly it belongs to $H$

Now using above stated result it is enough to show that $(i \ i+1) \in K \ \forall 1 \leq i<n $
$$(12...n)(12)(12...n)^{-1}=(23)$$ & $$(12...n)(23)(12...n)^{-1}=(34)$$ Proceeding inductively $$(12...n)^{i-1}(12)(12...n)^{-(i-1)}=(i \ i+1) \in K$$ $\blacksquare$

Keshav Agarwal
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  • You probably meant that $H$ is the subgroup generated by the set of adjacent transpositions. $H=\langle(12), (23),\ldots\rangle$ instead of $H={(12), (23),\ldots}$. – Nothing special Feb 14 '24 at 04:10