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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.

Shaun
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Valor Vaporeon
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    It is false in general that if $x=y^m$ then $y=x^{-m}$. Note that $x^{-m}=(y^m)^{-m} =y^{m(-m)}=y^{-m^2}$. This equals $y$ if and only if $-m^2\equiv 1\pmod{|y|}$. And you could have a cyclic group generated by $y$, with $x=y^2$, for example. – Arturo Magidin Jan 27 '24 at 22:43
  • I guess a "top-down" proof could be made using the fact that the only groups of order $22$ are $\Bbb Z_{22}$ and $D_{11}$. Use this to get to that point. This might be nuking it though. – Shaun Jan 27 '24 at 22:51
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    Do you know the formula for the size of the product of two subgroups? $|HK|=|H|,|K|/|H\cap K|$. – Arturo Magidin Jan 27 '24 at 23:02
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    It is false, in general, that the order of a product equals the lcm of the orders of the terms. – Gerry Myerson Jan 27 '24 at 23:14
  • If this weren't true, you could find two such elements $x,y$ that did not generate the group, so would be contained in a maximal subgroup of $G$. So for it to be true, every pair of such elements cannot be contained in any maximal subgroup. But then that means $\langle x\rangle$ must always be maximal. So you're attempting to prove: every cyclic subgroup is maximal. – Steve D Jan 28 '24 at 14:57

2 Answers2

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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...

calc ll
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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 $11^2>22$.

Alternatively, use . . .

Question on groups of order $pq$

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

Shaun
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