Generating $PSL(2,13)$ with two subgroups of coprime order Let $G$ be the group $G=PSL(2,13)$. I conjecture the following :
(1) If $H$ is a subgroup of order $3$ and $N$ is a subgroup of order $7$ (so that both $H$ and $N$ are cylic), then $H\cup N$ generates the whole of $G$.
(2) If $H$ is a subgroup of order $4$ and $N$ is a subgroup of order $13$ (so that $N$ is cylic but $H$ is not, as $G$ has no element of order $4$), then $H\cup N$ generates the whole of $G$.
Any ideas on how to show (1) or (2) ? I'd prefer computer-free proofs.
Context : If true, those conjectures would imply that $G$ has no subgroups of order $3\times 7$ or $4\times 13$, yielding a partial answer to this other question.
My thoughts: When $N$ or $H$ is cyclic with order different from the characteristic, we can generate it by an element that can be diagonalized in an algebraic closure of ${\mathbb F}_{13}$. I'm not sure how to continue from here though.
 A: Here's the outline of an answer to (2).
In $\text{SL}(2,13)$ we can see by inspection that 
$$
\left\{
\begin{bmatrix}
1 & x\\
0 & 1\\
\end{bmatrix}
\mid
x\in\mathbb{F}_{13}
\right\}
$$
is a Sylow-13-subgroup, and we can the calculate and see that
$$
\left\{
\begin{bmatrix}
\zeta & x\\
0 & \zeta^{-1}\\
\end{bmatrix}
\mid
x\in\mathbb{F}_{13}, \zeta\in\mathbb{F}_{13}^{*}
\right\}
$$
is its normaliser. 
Now in $\text{PSL}(2,13)$ let $H$ be of order $4$ and $N$ be of order $13$; we can take (by Sylow) $N$ to be the image of the matrix group above modulo $\{\pm I\}$. Let $K$ be a subgroup containing both $H$ and $N$; by Lagrange the order of $K$ is divisible by $4.13$. 
We want to exclude the possibilities $|K|/52=1,3,7$.
The first two are impossible; each would by Sylow have a normal Sylow-13-subgroup, and our calculation above shows that this is not so.
The third case is also impossible. If $|K|=4.13.7$ then its Sylow-7-subgroup must be normal. That is a cyclic subgroup of order 7 will be normalised by an element of order 13; as $13\not|\ (7-1)$ the element of 13 will centralise the element of order 7. That is, an element of order 13 is centralised by an element of order 7. Our calulation above shows that is not so. 
As to question (1)  you can check by Sylow that any proper subgroup of order divisible by 21 has a normal Sylow-7-subgroup. In each case this will then have to be normalised by an element of order $3$. But even in $\text{SL}(2,169)$ (where we can diagonalise an element of order $7$) we can check that the elements of order 7 are conjugate to their inverses only, so that the element of order 3 will centralise the element of order $7$. However, in $\text{SL}(2,13)$ an element of order 3 is conjugate to $\text{diag}(\omega, \omega^2)$ (where $\omega$ is the cube root of $1$ in $\mathbb{F}_{13}$), and we can calculate the centraliser and see that there are no elements of order $7$.  
