a question about abstract algebra, a question related to permutation

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Given $X=\{1,2,......n\}$, let us call a permutation $p$ of $X$ an adjacency if it is a transposition of the form $(i,i+1)$ for $i\le n-1$. Prove that $(i,j)$ is a product of an odd number of adjacencies.

I am stuck in this question, I try to use mathematical induction to prove this question. I know when $j=i+1$, it is obviously right. Then, I try to find how to construct a product of adjacencies which is identical to $(i,i+2)$, but I still have no idea how to construct it. I plan to use mathematical induction to prove it, but I am still not sure how to use it to prove it. Can someone help me solve this question, or can someone give me some hints? Thank you.

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Here's the full proof by induction:

We induct on the value of $|j-i|$. Clearly the property holds for $|j-i|$ = 1 because $(i,i \pm 1)$ is the product of one adjacency.

Now, assume it holds for $j-i \leq n$. Consider $(i,i+n+1)$. This equals $(i+n,i)(i+n,i+n+1)(i,i+n)$. By the inductive hypothesis, both $(i+n,i)$ and $(i,i+n)$ are expressible as the product of an odd number of adjacencies. An odd number plus one plus another odd number is odd, so $(i,i+n+1)$ can be expressed as the product of an odd number of adjacencies. Flipping the signs shows that $(i,i-n-i)$ is similarly expressible. This complete the induction.

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We can do it like this: (i,j)=(i,i+1)(i+1,i+2)......(j-2,j-1)(j-1,j)(J-1,j-2)......(i+2,i+1)(i+1,i) then I think it works(based on the hints above)