A group $G$ of order $22$ contains elements $x$ and $y$, where $x \neq 1$ and $y$ is not a power of $x$. Show that the subgroup generated by these two elements is the whole group.
I'm not allowed to use Sylow's theorems, only Lagrange.
Attempt: I can see that the order of $x$ must be either 2 or 11 (if it were 22 then $y$ would be a power of $x$ as $G=<x>$). –I feel that the order of $y$ also cannot be 22 but cannot prove it–. If the order of $x$ is 11 AND the order of $y$ is 11, then $|G|=21$ because $<x>\cap<y>=\varnothing$, a contradiction. This is where I am lost, because I don't see why the order of $x$ and $y$ cannot both be $2$. After this case, though, then I think I am done because the orders of $x$ and $y$ have to be either 11 and 2 or 2 and 11 (say the order of $x$ is 11). Then the set
$$ \{ e, x, x^2, \dotsc, x^{10}, y, yx, \dotsc, yx^{10} \} = G. $$
Best Answer
Let $H = \langle x,y\rangle$ the subgroup generated by $x$ and $y$. Since $x \neq 1$, it cannot be trivial, so it can have order $2,\, 11$ or $22$. If it had order $2$, then either $y = x$ or $y = 1$, so $y$ is a power of $x$, contradicting the premise. If the order were $11$, then $x$ is a generator of the cyclic group of prime order, hence $y$ is a power of $x$, contradicting the premise. So the only possibility that remains is $[H : 1] = 22$.