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Let $S_{n}$ be symmetric group. Then it is given by generators $\tau_{i}$ where $i=1,2,\ldots,n-1$ and relations $${\tau_{i}}^2$$ $$\tau_{i}\tau_{j}=\tau_{j}\tau_{i}\text{ for }i\neq j\pm1$$ $$\tau_{i}\tau_{i+1}\tau_{i}=\tau_{i+1}\tau_{i}\tau_{i+1}$$ I will be pleased if someone can present me detailed proof of this fact.

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    I would like to know if there is a conceptually different (yet reasonable) proof besides the reflection group approach. So, I am offering a bounty on this question.2012-11-29
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    Please note that I think that the current answer *does* contain enough detail to be an answer to the original question, but I didn't want to start a new question with basically the same content, and this was the closest one in the choices for bounty messages.2012-11-29
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    In the book "Braid groups" by Kassel and Turaev, there is an elementary proof in chapter 4.1. http://books.google.de/books?id=y6Cox3XjdroC&pg=PA151&lpg=PA151&dq=generators+and+relations+of+the+symmetric+group&source=bl&ots=eB8Uezuoft&sig=DJUTs9DVO0ikhRXwu_I5iy8e1ck&hl=en&sa=X&ei=L6i3UOfqKIPJtAaD-4C4CA&ved=0CDsQ6AEwAzgK#v=onepage&q=generators%20and%20relations%20of%20the%20symmetric%20group&f=false2012-11-29
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    If you don’t want the reflection group aspect, you can do it completely combinatiorally, using parabolic subgroups in Coxeter groups. It is however a “reflection group approach in disguise”, so I don’t know if this is what you’re looking for.2012-11-29
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    @EwanDelanoy Basically, I am looking for an answer that can easily be given to students who have just learned the basics about groups (such an answer is now here), and I am also very interested in other elegant answers of any theoretical level that use some neat trick I am unaware of. So, I am not really looking for a way to "disguise" the advanced aspect.2012-11-29
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    @Phira : I see. I think you'll agree YACP's solution below is the perfect solution, then.2012-11-30

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