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Does $\langle ( 1, 3), (1 ,2 ..., 10)\rangle $ generate the group $S_{10}$ ?

I think that's it doesn't because every use of $(1, 3)$ makes a "jump" between at least two numbers. So we can get for example to $(1 ,2)$. However I can't prove it formally.

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    $\begingroup$ HINT: Think of all the odds at once, against all the evens at once. $\endgroup$ – Empy2 Apr 22 '19 at 13:52
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    $\begingroup$ What you need is the difference $3-1$ in your "jump", which should be coprime to $10$, but isn't. $\endgroup$ – Dietrich Burde Apr 22 '19 at 13:59
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By definition $\langle (1,3),(1,2,...,10)\rangle$ is the smallest subgroup of $S_{10}$ which contains the elements $(1,3),(1,2,...,10)$. So let's define:

$H=\{\sigma\in S_{10}: i\equiv j \pmod{2} \iff \sigma(i)\equiv\sigma(j) \pmod {2}\ \forall 1\leq i,j\leq 10\}$

Now check that $H$ is a proper subgroup of $S_{10}$ which contains both permutations $(1,3)$ and $(1,2,...,10)$. Hence $S_{10}$ is not the smallest group which contains both permutations $(1,3),(1,2,...,10)$.

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It is well-known that $(123\dots n)$ and $(ab)$ generate $S_n$ if and only if $\text{gcd}(|a-b|,n)=1$. For $n=10$ and $(ab)=(13)$ this greatest common divisor is different from $1$.

References:

How does $(12\cdots n)$ and $(ab)$ generate $S_n$?

Necessary and Sufficient conditions to generate $S_n$

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