A concrete example of involution in group theory

Question

I am reading the textbook "Introduction to Modern Algebra, Joyce 2017" and in the Cyclic groups and subgroups section, there is a following sentence about involution.

An involution $a$ is an element of a group which is its own inverse, $a^{-1} = a$. Clearly, the order of an involution $a$ is $2$ unless $a = 1$, in which case the order of $a$ is $1$.

I was trying to come up with a concrete example, but I am having difficulties. Maybe when $a = \frac{1}{2}, \frac{1}{2}^2 = 1$? I also searched the wikipedia page for a concrete example, but wasn't able to find anything.

Can someone provide an example with numbers for me please? I want to understand why the author writes "clearly" that the order of an involution $a$ is $2$. Thank you

Answer 1

Example: in the cyclic group $\mathbb{Z}_{12}$, $6$ is an involution since $6+6=0$ in $\mathbb{Z}_{12}$. More generally, in $\mathbb{Z}_{2n}$, the element $n$ is an involution.

Notice that the requirement $a=a^{-1}$ is equivalent to $aa=1$ (just multiply both sides by $a$). Therefore an involution must satisfy $a^2=1$. So the only way it does not have order $2$ is if it has order smaller than $2$, i.e., $a=1$.

Answer 2

An involution in a group is any non-trivial element of order two (not a subset of them, as the phrase you mention suggests, and also the identity is explicitly omitted). That is, $a$ is an involution if and only if $a\neq1$ and $a^2=1$: $$ \begin{align*} a&=a^{-1}\\ \Leftrightarrow a\cdot a&=a\cdot a^{-1}&\text{(left multiply by $a$)}\\ \Leftrightarrow a^2&=1 \end{align*} $$ For a concrete example of an involution, consider the set $\{0, 1\}$ under addition modulo $2$. Here, the identity is $0$, and $1+1=2=0\pmod2$.

Answer 3

An involution is, in essence, anything that undoes itself. For example, flipping a coin over by one axis, the action of turning a key for certain padlocks, or adding one modulo two.

Answer 4

Take the general linear group $G=GL_n(K)$ and the element $A=-I_n$. It has order $2$ because of $A^2=I_n$ and hence is an involution.