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I am self-studying group theory, and find it frustrating that problems on my books do not have solutions provided, here is one:

Show that the symmetry group $S_n$ can be generated by the set $\{(1 2),(1 2 \cdots n)\}$ of a transposition and a n-cycle

My thoughts:

To prove that elements in this set actually generate $S_n$, I shall show that every permutation $\sigma \in S_n$ can be written as a combination of $(1 \ 2)$ and $(1 \ 2 \ \cdots \ n)$. How shall I proceed to solve this problem?

user26857
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Rescy_
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    There might be a simpler approach, but I had this as a homework exercise in one of my Algebra courses, and my approach then was to show that you can use the transposition and the n-cycle to produce any transposition, which would solve the problem since $S_n$ is generated by transpositions. I did this by first showing that the group generated contains the transpositions (1 k) for all k, from there that it contains (k k+1), and then finally that it contains arbitrary transpositions. – Dylan Nov 22 '15 at 00:09
  • Show transpositions generate it, and that in fact transpositions of the form $(1i)$ generate it. Now conjugate $(12)$ by such cycle. – Pedro Nov 22 '15 at 00:10
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    I think I might have had it the wrong way round. I might first have shown that you can get (k k+1) and used that to show you can get (1 k). – Dylan Nov 22 '15 at 00:13
  • As you progress into higher mathematics you should expect to spend more and more time on a single problem. You'll have no idea how to solve it at first. You spend a few hours of intense concentration, flipping around in the book (best to look only in the book the problem is in), and you eventually think of something that might work. Then you see if it works. Etc. Looking at solutions to these problems will be little help in solving new problems because the new problems will be completely different. – Matt Samuel Nov 22 '15 at 01:14

1 Answers1

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Filling in the details for my comment:

Let $\sigma = (1\;2\;3\cdots n)$ and $\tau=(1\;2)$. We will first show that using $\sigma$ and $\tau$, we can generate all transpositions of the form $(k\enspace k+1)$. The main idea is that we can use the cycle $\sigma^{-1}$ to "shift up" the numbers which we are permuting so that the pair that we want to transpose end up in the $1$ and $2$ places, use the transposition $\tau$ to swap these two elements, and then use $\sigma$ to "shift everything back".

More explicitly, we can generate the transposition $(k\enspace k+1)$ as follows:

  • We shift the numbers up $k-1$ spaces so that $k$ occupies the position of $1$. This is accomplished by applying the permutation $\sigma^{-(k-1)}$.
  • We swap the numbers in the first and second position (now equal to $k$ and $k+1$) by applying the transposition $\tau$.
  • We move the numbers back up by $k-1$ positions so that all of the numbers except $k$ and $k+1$ are in their original places. This is accomplished by applying the permutation $\sigma^{k-1}$.

The above argument leads us to expect that $$ (k\enspace k+1) = \sigma^{k-1} \tau \sigma^{1-k} $$

Alternatively, we could show that we can generate the transposition $(k\enspace k+1)$ by building up the set of transpositions inductively. Specifically, we already know that we can generate $(1\; 2)$ since it is just equal to $\tau$. We can then show that if we can generate the transposition $(k\enspace k+1)$, then we can also generate the transposition $(k+1\enspace k+2)$. The key step here is to use the identity $$ (k+1\enspace k+2) = \sigma (k\enspace k+1) \sigma^{-1} $$

We now show that we can generate all transpositions of the form $(1\; k)$ for $k$ from $1$ to $n$. We can build up this set of transpositions inductively too. We know, as the base case, that we can generate the transposition $(1\; 2)$ since it is just equal to $\tau$. Now suppose that we know that we can generate $(1\; k)$ for some $k<n$. Then we can also generate $(1\enspace k+1)$ since we have that $$(1\; k+1) = (1\;k)(k\enspace k+1)(1\;k) $$ and we know already that we can generate the transpositions $(1\; k)$ and $(k\enspace k+1)$.

Finally, we can show that (using $\sigma$ and $\tau$) we can generate arbitrary transpositions. If $(m\; k)$ is any transposition, then we have that $$ (m\; k) = (1\; m)(1\; k)(1\; m)$$

Thus the group generated by $\sigma$ and $\tau$ includes all transpositions. Since we know that $S_n$ itself is generated by transpositions, we see that $\sigma$ and $\tau$ must in fact generate all of $S_n$.

Dylan
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