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Prove that every permutation in $S_k$ is the product of transpositions of the form $(j, j + 1).$

I proved the case $n=2$ for my base case... so $(12)=(21)$ and $(21)=(12)(12)$ then I proved $n=3$ and found that that $(123)=(12)(12)(23)(23)$ etc... (I don't want to type them all). Anyways, I was going to say for the inductive step that since the only additional term needed between $n=2$ and $n=3 $is$ (2,3)$ that I can say the only term between $n=k$ and $n=k+1$ that is needed is $(k,k+1)$ but I feel this has a lot of holes. How else could I do this inductively?

user26857
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MathIsHard
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1 Answers1

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There seems to be some confusion in the question about the roles of $n$, $k$ and $j$. I'll denote the number of elements being permuted by $n$.

Note that any transposition $(i,j)$ (with $i\lt j$) can be written as a product of adjacent transpositions in the form $(i,i+1)(i+1,i+2)\cdots(j-1,j)\cdots(i+1,i+2)(i,i+1)$. Thus it suffices to show that any permutation can be written as a product of transpositions.

Given a permutation $\sigma$ that maps $n$ to $i$, the transposition $\tau=(i,n)$ maps it back to $n$, so $\pi=\tau\sigma$ maps $n$ to $n$ and thus only permutes the other $n-1$ elements among themselves. By the induction assumption, this permutation of $n-1$ elements can be written as a product of transpositions, and thus so can $\sigma=\tau\pi$.

joriki
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