Bijective functions of order $2$ that generate infinite subset

Question

Let $X$ be a non-empty, possibly infinite set and let $\Sigma_X$ be the set of all bijections $f: X \to X$. Prove that if $X = \mathbb{Z}$ there are elements $\sigma, \tau \in \Sigma_X$, each of order $2$, that generate a subgroup of infinite order.

I don't understand what this is asking to prove since if $\sigma$ and $\tau$ are bijective functions of order $2$, that means that they are their own inverses. But the only such bijective function is the identity function, so how can there be more than one such bijection of order $2$ in the first place?

Answer 1

Functions $f : X \to X$ that satisfy $f\circ f = \mathrm{id}_X$ are called involutions.

• Notice that if $f : X \to X$ is a bijection, then we can partition $X$ according to the orbits of $f$, i.e., sets of the form $$ \{ f^{\circ n}(x) : n \in \Bbb{Z} \}. $$ If $f$ is an involution, then the orbits are either a singleton $\{x\}$ when $f(x) = x$ or of the form $\{x, y\}$ when $y = f(x) \neq x$.

• Conversely, for any partition $P = \{A_i : i \in I\}$ of $X$ into sets of size 1 or 2 (i.e. $|A_i| = 1$ or $2$ for all $i \in I$) gives rise to an involution which swaps elements in each $A_i$.

Now consider two involutions $\sigma, \tau$ on $\Bbb{Z}$ defined by

$$ \sigma(x) = \begin{cases} x+1 & \text{if } x \text{ is even} \\ x-1 & \text{if } x \text{ is odd} \end{cases}, \qquad \tau(x) = \begin{cases} x-1 & \text{if } x \text{ is even} \\ x+1 & \text{if } x \text{ is odd} \end{cases}. $$

That is, $\sigma$ corresponds to the partition $\{\cdots, \{-2, -1\}, \{0, 1\}, \{2, 3\}, \cdots \}$ and $\tau$ corresponds to the partition $\{\{-1, 0\}, \{1, 2\}, \{3, 4\}, \cdots \}$. Consequently

$$ \sigma(0) = 1, \quad \sigma(2) = 3, \quad \sigma(4) = 5, \quad \cdots $$

and

$$ \tau(1) = 2, \quad \tau(3) = 4, \quad \tau(5) = 6, \quad \cdots. $$

From this observation, we find that

$$(\tau\circ\sigma)(0) = 2, \quad (\sigma\circ\tau\circ\sigma)(0) = 3, \quad (\tau\circ\sigma\circ\tau\circ\sigma)(0) = 4, \cdots $$

hence $\sigma$ and $\tau$ generates infinite subgroup.

Answer 2

Consider the following elements in $\Sigma_X$ where $X = \mathbb{Z}$ $$ \sigma(i) = \begin{cases} i &: i \leq 0\\ i+1 &: i \in \{1,3,5,\ldots\} \\ i-1 &: i\in \{2,4,6,\ldots\} \end{cases} $$ $$ \tau(j) = \begin{cases} i &: i\leq -1 \\ i-1 &: i\in \{1,3,5,\ldots\} \\ i+1 &: \{0,2,4,\ldots\} \end{cases} $$ Then $o(\sigma) = o(\tau) = 2$, but $\sigma\tau$ has infinite order (look at the image of $2$ under $(\sigma\tau)^n$)