Invertible Elements in Finite Groups that are NOT Equal to Themselves

Consider a finite group $(G, \cdot)$. For each $x \in G$ let $-x \in G$ denote the inverse of $x$ under the operation $\cdot$. Now consider the subset $S \subseteq G$ of invertible elements whose inverse is not equal to themselves, that is:

(1)

\begin{align} \quad S = \{ x \in G : x \neq -x \} \end{align}

We will now look at some interesting results regarding the number of elements in $S$.

Proposition 1: If $(G, \cdot)$ is a finite group, $e \in G$ is the identity with respect to $\cdot$, and $S \subseteq G$ where $S = \{ x \in G : x \neq - x \}$ then $\mid S \mid$ is even.

• Proof: The subset $S \subseteq$ is the set of all invertible elements whose inverse is not equal to themselves. Hence, for each $x \in S$ there exists a $-x \in S$ such that $x + (-x) = e$ and where $x \neq -x$. Conversely, for $-x \in S$ we have that $x \in S$ is such that $(-x) + x = e$. Therefore the elements in $S$ come pairs from the uniqueness of inverses under the operation $\cdot$, so $\mid S \mid$ is even.

Corollary 2: Let $(G, \cdot)$ be a finite group and let $e \in G$ be the identity with respect to $\cdot$. If $\mid G \mid$ is odd then the number of elements in $G$ that are equal to their inverse with respect to $\cdot$ is odd, and if $\mid G \mid$ is even then the number of elements in $G$ that are equal to their inverse with respect to $\cdot$ is even.

• Proof: Let $S = \{ x \in G : x \neq -x \}$. From Proposition 1 we have that $\mid S \mid$ is even. Now, consider the set of elements in $G$ that are equal to their inverse with respect to $\cdot$. This set can be written in the form:

(2)

\begin{align} \quad G \setminus S = \{ x \in G : x = -x \} \end{align}

• Clearly $G = S \cup (G \setminus S )$ and $S \cap (G \setminus S) = \emptyset$ and since $G$ is a finite set, we have that:

(3)

\begin{align} \quad \mid S \mid + \mid G \setminus S \mid &= \mid G \mid \\ \quad \mid G \setminus S \mid &= \mid G \mid - \mid S \mid \end{align}

• Suppose that $\mid G \mid$ is odd. Then $\mid G \mid + \mid S \mid$ is odd, so $\mid G \setminus S \mid$ is odd, that is, the number of elements in $G$ that are equal to their inverse with respect to $\cdot$ is odd. Similarly, suppose that $\mid G \mid$ is even. Then $\mid G \mid + \mid S \mid$ is even, so $\mid G \setminus S \mid$ is even and so the number of elements in $G$ that are equal to their inverse with respect to $\cdot$ is even. $\blacksquare$

Corollary 3: Let $(G, \cdot)$ be a finite group and let $e \in G$ be the identity with respect to $\cdot$. If $\mid G \mid$ is even then there exists an element $x \in G$ such that $x = x^{-1}$ and $x \neq e$.

• Proof: Suppose that $\mid G \mid$ is even. By Corollary 2 above, this implies that $\mid G \setminus S \mid$ is even too, that is, the number of elements $x \in G$ that are equal to their inverse with respect to $\cdot$ is even. We see that $e \in G \setminus S$ since $e \cdot e = e$. Since $\mid G \setminus S \mid$ is even, there must then exist another element $x \in G \setminus S$ such that $x = x^{-1}$. $\blacksquare$

It is important to note that Corollary 2 need not hold if $\mid G \mid$ is odd. For example, consider the group $(\mathbb{Z}_3, \cdot)$ of integers modulo $3$ where for $x, y \in \mathbb{Z}_3$ we define $x \cdot y = (x + y) \mod 3$. We noted earlier that for each $x \in \mathbb{Z}_3$ that the inverse of $x$ will be $3 - x$ if $x \neq 0$ and $0$ if $x = 0$. We have that $e = 0$, so clearly $e \cdot e = 0 = e$. Furthermore, $1 \cdot 2 = 0 = e$. So as you can see, there does not exist an element $x \in \mathbb{Z}_3$ such that $x \neq e$ and for which $x = -x$.