# A concrete example of involution in group theory

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

• @DietrichBurde So far just cyclic groups...I just started learning about groups in general. – Evan Kim Apr 4 at 19:46
• Well, matrix groups already appear in linear algebra. They are good examples, too. – Dietrich Burde Apr 4 at 21:07
• From Wolfram Mathworld: "Group Involution: An element of order 2 in a group (i.e., an element $A$ of a group such that $A^2=I$, where $I$ is the identity element." That's two inequivalent definitions! I think in group theory it virtually always means an element of order $2$, but an involutory function is allowed to be the identity. – Derek Holt Apr 5 at 1:07

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., $$a=1$$.

• ah, good observation. Thanks for pointing that out. That makes a lot more sense – Evan Kim Apr 4 at 16:00
• When group theorists talk about involutions they generally mean elements of order $2$ - not the identity element. – Derek Holt Apr 5 at 1:10

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.

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.

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 $$1+1=2=0\pmod2$$.

• Your "That is" sentence does not agree with your first sentence. The identity satisfies $a^2=1$, but does not have order $2$. (Anyway all elements of order $2$ are non-trivial.) – Derek Holt Apr 5 at 3:44
• @DerekHolt Sorry, silly omission! I've corrected my answer. – user1729 Apr 5 at 12:15