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Need which property of algebra, or otherwise; to prove it.

jiten
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2 Answers2

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Hint $S_n=\langle (12),(12\dots n)\rangle $

Proof:

It suffices to show you can get any transposition ($2$-cycle).

Note $(12\dots n)^{k}(12)(12\dots n)^{-k}=(k+1 k+2)$.

From there you can get the rest: $(12)(23)(12)=(13)$ etc.

calc ll
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  • But, then need still proof that : $(12\dots n)^{-k}=((12\dots n)^k)^{-1} =((12\dots n)^{-1})^k$. – jiten Jul 05 '22 at 13:44
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    In any group, $g^{mn}=(g^m)^n=(g^n)^m$. To prove it rigorously takes a little work. The special case you are asking for follows from uniqueness of inverses and $g^{m+n}=g^mg^n$. – calc ll Jul 05 '22 at 14:19
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Let $s_i=(i,i+1)$ where $1\leq i$ and we think this definition$\mod n$, zero being $n$. Let $\alpha=(1,2,3,...,n)$. Then it can be checked that $$\alpha s_i\alpha^{-1}=s_{i+1}.$$ Hence, all $s_i$s are generated by $s_1$ and $\alpha$.

But on the other hand, we have the identities $s_j(i,j)s_j=(i,j+1)$ which implies that all 2-cycles are generated by $s_i$s. Hence, all 2-cycles are generated by $s_1$ and $\alpha$.

But, we know that the symmetric group is generated by 2-cycles.

Hence, the group $S_n$ is generated by $s_1$ and $\alpha$.

Bob Dobbs
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