Can a group of order $55$ have exactly $20$ elements of order $11$? (Clarification)

By Lagrange Theorem, any element has order $1,5,11$ or $55$.

Case 1 There is an element of order $55$, then group is cyclic (thus isomorphic to $\mathbb{Z}/55\mathbb{Z})$ , and it is easy to check that there are exactly 10 elements of order 11.

Case 2 There is no element of order $55$. Then any element has order $1,5$ or $11$.

There are 55 elements in total. 1 has order 1, 20 have order 11 and the remaining $55-20-1=34$ have order 5.

Now, each element $x$ of order $5$ generates the following subgroup: $\langle x,x^2,x^3,x^4, e \rangle$, which contains four elements of order $5$.

If you pick another element $y$ of order $5$ then either $\langle y\rangle=\langle x\rangle$ or $\langle y\rangle \cap \langle x\rangle =\{e\}$.

This shows that the elements of order $5$ can be grouped in disjoint groups of 4 elements. Thus, their number is a multiple of $4$.