Can a group of order $55$ have exactly $20$ elements of order $11$? (Clarification) After I asked this question, which I now understand, I came across a similar question. But I don't understand the answer that was chosen; particularly the part about $34$ elements of order $5$ (but the rest of the answer isn't clear either). Could someone please clarify this for me? Thank you.
 A: 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 $4$.
A: Let $G$  be a group of order 55.
Suppose $G$ contains exactly $20$ elements of order $11$.
Suppose $a \in G$ and $|a|=11$ 
then $⟨a⟩$ is a cyclic subgroup of $G$ with order $11$ and contains exactly $10$ elements of $11$ order.
Now choose $b$ belonging to $G$ such that $b$ is not in $⟨a⟩$ and $|b|=11$ 
[$b$ is taken from the rest $10$ elements of order $11$] 
then ⟨b⟩ is a subgroup of order 11.
Now, $|⟨a⟩ \cap ⟨b⟩|$ can be $1$ or $11$.
If $|⟨a⟩ \cap ⟨b⟩| = 11$ then, $$|⟨a⟩ \cap ⟨b⟩|=|⟨a⟩|=|⟨b⟩|$$ but $⟨a⟩$ is not equal to $⟨b⟩$ by construction.
Hence, $|⟨a⟩ \cap ⟨b⟩| = 1$,  which implies that $$|⟨a⟩ . ⟨b⟩|=|a||b|=11 \times 11=121 > (55=|G|)$$  which is a contradiction.
