Find the order of a subgroup of $S_5$ generated by two elements The Question I am trying to solve reads :
Find the order of a subgroup of $S_5$ generated by the elements $s=(123)$ and $r=(12345)$. Is this group sovable ? Is it nilpotent ?
I tried to compute a couple of terms but then I realized this subgroup is way large, and computing the elements one by one might not be a good idea (Is there another way to compute the order ?)
My calculation shows : $r, r^2, r^3, r^4, rs, r^3s, sr $ are all $5-$cycles but there are elements like $r^2s=(14)(25)$ and also $r^4s=(354)$...It seems that all sorts of things can happen in this subgroup...I do not even know how to approach the solvability and nilpotency before I know what the subgroup is...
I already appreciate your hints/ideas/answers 
 A: well you can see that your subgroup, say $H$ has $(123),(12345),(123)(12345)=(13452),(123)(13452)=(345), (123)(345)=(12345)$ etc , so as Robert said order of your subgroup is a multiple of $15$, say, $15k$ and will divide $120$, so $k$ can have values $2,4,8$, i.e order of subgroup can be $30,60$ or $120$, now check that is it equal to $S_5$ or $A_5$? 
Write an arbitrary element of your subgroup, it will be like $\Pi(123)^i(12345)^j$ where $i,j \in $ {$1,0,-1$}, so you will never have any odd permutation in it, so it is not $S_5$. So it only has even permutations, so it can be of order $30$ or $A_5$.
Now your subgroup has $(123)$ and $(345)$, so it will also have the conjugate $(345)^{-1}(123)(345)=(124)$. 
Now consider subgroup $A=<(123)(345)>$ of your subgroup $H$, You note that, $A$ acts transitively on {$1,2,3,4,5$}, as for any two elements $i$ and $j$ in {$1,2,3,4,5$} there exist an element in $A$, say $\sigma$ s.t. $\sigma(i)=j$, It is easy to see, right? (use the fact that both generating elements of $A$ do not fix a common element of {$1,2,3,4,5$} )? 
Now consider another subgroup of $H$, say $B=<(123),(124)>$, as both generating elements of $B$ are not moving $5$, this $B$ is a permutation group on {$1,2,3,4$}, and again by the fact that both generating elements of $B$ do not fix a common element of {$1,2,3,4$}, it acts transitively on the set {$1,2,3,4$}.
Now consider third subgroup $C=<(123)>$ of $H$,  it acts transitively on {$1,2,3$} (obviously).
Now is the time to use $\textbf{Orbit stabiliser theorem}$ on all three $A,B$ & $C$. as $C$ acting transitively on set  of $3$ elements (i.e. orbit has order $3$)  gives you |$C|=3k$ now similarly order of $B$ is a multiple 0f $4$ and order of $A$ is a multiple of $5$ abut notice that $C<B<A<H$, which implies that |$A$| is a multiple of $3,4$ and $5$, i.e. a multiple of $60$ and $A<H$ implies |$H$| is also a multiple of $60$, so order $30$ ruled out, so $60$ or $120$ , but 120 was ruled out before, so |$H=60$|, and therefore $H\cong A_5$.
Now $A_5$ is not solvable, (because if $A_5$ is solvable then $\mathbb{Z_2}$ being solvable, will imply $S_5$ is solvable, which you must know is not, as quintics are not solvable) and hence not nilpotent as nilpotent implies solvable.
