Find an infinite group that has exactly two elements with order $4$?
Let $G$ be an infinite group for all $R_5$ (multiplication $\mod 5$) within an interval $[1,7)$. So $|2|=|3|=4$. Any other suggestions, please?
Find an infinite group that has exactly two elements with order $4$?
Let $G$ be an infinite group for all $R_5$ (multiplication $\mod 5$) within an interval $[1,7)$. So $|2|=|3|=4$. Any other suggestions, please?
The one we bump into most often is the multiplicative group of non-zero complex numbers. There are all sorts of minor variants of the idea, such as the complex numbers of norm $1$ under multiplication. One can disguise these groups as matrix groups, or geometric transformation groups.
Pick $G$ an infinite group with no torsion (e.g. $\mathbb Z$ or $\mathbb Q$), then $\mathbb Z / 4 \mathbb Z \times G$ works.
As a non-abelian example, you can also take the free product $\mathbb Z / 4 \mathbb Z * G$.
Cyclic group generated by $1$ and $I =<1,i>.$