Determine every group with order 22

I have the following theorem:

Let $G$ a group with $|G| = pq$ , with $p,q$ primes such that $p<q$ and $p$ $| (q-1)$. Then $G \approx \Bbb Z_{pq}$

In this case we got $|G| = 2 · 11 = 22$ with $2$ $| (11-1)$ so $G \approx \Bbb Z_{22}$ for every group with 22 elements.

But I also know $G \approx D_{11}$ (the diedric group) because $|D_{11}| = 22$. And $\Bbb Z_{22} \not \approx D_{11}$

What am I missing here?

The "theorem" is only true with the hypothesis that $p$ does not divide $q-1$.

In this case, the proof goes by checking that there is exactly one Sylow $p$-subgroup and one Sylow $q$-subgroup, in which case both are normal and the group is the internal direct product $C_p \times C_q$.

It is just the other way around. If $p$ does not divide $q-1$, then we only have the identity homomorphism $\Phi:C_q\to\mathrm{Aut}(C_p)\cong C_{p-1}$, i.e., only one possible group of order $pq$, the cyclic one. For $p\mid q-1$ we have two possibilities:

Structure of groups of order $pq$, where $p,q$ are distinct primes.

So there are exactly two groups of order $22$ up to isomorphism and one is abelian, the other not.