The solvability of a subgroup in $S_{11}$

Question

Is the subgroup in $S_{11}$ generated by cycles $(1, 2, 3, 4, 5)$ and $(1, 6, 7, 8, 9, 10, 11)$ solvable?

The cycles generating our subgroup have lengths $5$ and $7$, if multiplied, we get a cycle of length $11$, I believe that if we continue multiplying, we can come to the conclusion that these cycles generate $A_{11}$, but it seems that if this is true, it can be done in a more elegant way

Answer 1

Let $a=(1,2,3,4,5)$, $b=(1,6,7,8,9,10,11)$ and $G=\langle(a,b)\rangle$ To prove that $G=A_{11}$ we can proceed as follows. We multiply permutations from left to right, e.g. $(1,2)(1,3)=(1,2,3)$.

1. $b^{-1}ab=(6,2,3,4,5)$. It follows that $(i,2,3,4,5)\in G$ for any $i\in\overline{6,11}$.

2. $(i,2,3,4,5)(j,5,4,3,2)=(i,j,5)$, $i,j\in\overline{6,11}$. It follows that $(i,j,k)\in G$ for any $i,j\in\overline{6,11}$ and $k\in\overline{1,5}$.

3. Similarly, $(i,6,7,8,9,10,11)\in G$ for any $i\in\overline{1,5}$ and

4. $(i,j,k)\in G$ for any $i,j\in\overline{1,5}$ and $k\in\overline{6,11}$.

The remaining cycles of length $3$ are obtained from the ones mentioned in (2) and (4). For example $(1,2,3)=(1,2,6)(3,7,6)(1,6,7)$

We see that the group $G$ contains all cycles of length $3$, so $G=A_{11}$.

I don't know how elegant this way is.

Answer 2

The problem was to decide whether the group is solvable. It is clearly transitive hence primitive, since $11$ is prime. If it was solvable then a minimal normal subgroup would be transitive and abelian, and hence of order $11$. But the normalizer of a subgroup of order $p$ in $S_p$ is easily seen to be a Frobenius group of order $p(p-1)$. In particular, it does not include $5$- and $7$-cycles when $p=11$, so we have a contradiction, and the group is not solvable.