Group of order $22$ is generated by two elements

Question

I am trying to solve the below problem. I think this problem has been addressed on this website before, but I'd really like to work through the details myself without reading a prepared solution.

Let $G$ be a group of order $22$ and $x,y$ elements of $G$ where $x \neq e$ and $y \neq x^n$ for any $n$. Prove that $G$ is generated by $x$ and $y$.

The first thing I thought to do was use Lagrange's theorem. Any element of $G$ must have order dividing $|G| = 22$. The possibilities are $1,2,11,22$. As $x \neq e$, $|x| \neq 1$. As $y$ is not equal to a power of $x$, $G$ cannot be a cyclic group of order $22$ generated by $x$, so $|x| \neq 22$. This leaves only $|x| = 2$ or $|x| = 11$. It also means that $y \neq e$, or else there would be a power of $x$ which is equal to $y$ (since $x$ has finite order). If $G$ were cyclic generated by $y$, then $x = y^m$ for some $m$ so $y = x^{-m}$, which is not allowed, so $y$ also does not have order $22$. So both $x$ and $y$ have order $2$ or order $11$.

I have also previously proved that a group of even order has an element of order $2$. If both $x$ and $y$ had order $11$, I don't think this would be possible. The product $xy$ has order equal to (if I'm not mistaken) $\text{lcm}(x,y)$, which has to be at least $11$. This is circular reasoning because I'm presupposing that $x$ and $y$ are the generators of the group, but I think I can say that if $\langle x, y \rangle$ is the group generated by $x$ and $y$, it certainly cannot have an element of order $2$ if $x$ and $y$ both have order $11$. I think I would also run out of elements if both $x$ and $y$ had order $2$. The main group I know of elements of order $2$ is the Klein-four group, which is generated by two elements of order $2$ (and is isomorphic to $\mathbb{Z}/2 \times \mathbb{Z}/2$. So my sense is that one of $x$ or $y$ must have order $11$ and the other order $2$.

I'm a bit stuck on this point, but would appreciate some direction on how to proceed.

Answer 1

Hint: Use $y\notin\langle x\rangle$ to consider $\langle y\rangle \cap\langle x\rangle =\{e\}$. What possible orders could $x,y$ each have? Use Lagrange's Theorem and the fact that

$$|HK|\cdot |H\cap K|=|H|\cdot \lvert K\rvert $$

for any $H,K\le G$ and any group $G$.

Also $4\nmid 22$ and $11^2>22$.

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Alternatively, use . . .

Question on groups of order $pq$

. . . to determine the group is $D_{11}$ (because it is not cyclic) and proceed from there.

Answer 2

Hint: Say the elements $x,y$ are of orders $2,11$.

Then consider $\langle x,y\rangle $.

You might also have two elements of order $2$, but you can argue the same way...