# Does $\langle ( 1 ,3), (1, 2 … ,10)\rangle$ generate the group $S_{10}?$ [duplicate]

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.

## marked as duplicate by Derek Holt group-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 22 at 14:13

• HINT: Think of all the odds at once, against all the evens at once. – Empy2 Apr 22 at 13:52
• What you need is the difference $3-1$ in your "jump", which should be coprime to $10$, but isn't. – Dietrich Burde Apr 22 at 13:59

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 $$(1,3),(1,2,...,10)$$.

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 $S_n$