Functions from a Group To Itself

# Functions from a Group to Itself

Recall that a function $f : A \to B$ is a rule that assigns to each element $x \in A$ exactly one element $y \in B$ which we denote by $f(x) = y$. On the Injective, Surjective, and Bijective Functions page we also defined what it meant for a function to be injective, surjective, or bijective.

Now consider a group $(G, \cdot)$. Since $G$ itself is a set, we can define a variety of functions $f : G \to G$. We will now look at some functions that can be formed and look at their various properties.

 Proposition 1: Let $(G, \cdot)$ be a group and let $f : G \to G$ be defined for each $x \in G$ by $f(x) = x^{-1}$. Then $f$ is bijective.
• Proof: To show that $f$ is bijective we must show that $f$ is both injective and surjective.
• Suppose that $f(x) = f(y)$. Then:
(1)
\begin{align} \quad x^{-1} = y^{-1} \\ \quad x \cdot x^{-1} = x \cdot y^{-1} \\ \quad e = x \cdot y^{-1} \end{align}
• Since $x \cdot y^{-1} = e$ then by the uniqueness of the inverses with respect to $\cdot$ we have that $x = y$, so $f$ is injective.
• Now let $y \in G$. Then $y^{-1}$ is such that:
(2)
\begin{align} \quad f(y^{-1}) = (y^{-1})^{-1} = y \end{align}
• So for every $y \in G$ there exists $y^{-1} \in G$ such that $f(y^{-1}) = y$. Therefore $f$ is surjective.
• Since $f$ is both injective and surjective we have by definition that $f$ is bijective. $\blacksquare$
 Proposition 2: Let $(G, \cdot)$ be a group and let $f : G \to G$ be defined for each $x \in G$ and some $a \in G$ by $f(x) = a \cdot x$. Then $f$ is bijective.
• Proof: Once again we will first show that $f$ is injective and then that $f$ is bijective. Suppose that $f(x) = f(y)$. Then:
(3)
\begin{align} \quad a \cdot x = a \cdot y \\ \quad a^{-1} \cdot (a \cdot x) = a^{-1} \cdot ( a \cdot y) \\ \quad (a^{-1} \cdot a) \cdot x = (a^{-1} \cdot a) \cdot y \\ \quad e \cdot x = e \cdot y \\ \quad x = y \end{align}
• Therefore whenever $f(x) = f(y)$ we have that $x = y$, so $f$ is injective.
• Now since $a \in G$ and since $(G, \cdot)$ is a group we must have that $a^{-1} \in G$, and so for $(a^{-1} \cdot x) \in G$ (from the closure property of $\cdot$ on $G$) we have that:
(4)
\begin{align} \quad f(a^{-1} \cdot x) = a \cdot (a^{-1} \cdot x) = (a \cdot a^{-1}) \cdot x = e \cdot x = x \end{align}
• Therefore for every $x \in G$ there exists a $y = a^{-1} \cdot x \in G$ such that $f(y) = f(a^{-1} \cdot x) = x$, so $f$ is surjective.
• Since $f$ is both injective and surjective then by definition $f$ is bijective. $\blacksquare$
 Corollary 3: Let $(G, \cdot)$ be a group and let $f : G \to G$ be defined for each $x \in G$ and some $a \in G$ by $f(x) = a \cdot x$. Then $f^{-1}(x) = a^{-1} \cdot x$.
• Proof: From Proposition 1 we have that $f$ is bijective and hence $f^{-1} : G \to G$ exists. If we let $g : G \to G$ be defined by $g(x) = a^{-1} \cdot x$ then:
(5)
\begin{align} \quad f(g(x)) = f(a^{-1} \cdot x) = a \cdot (a^{-1} \cdot x) = (a \cdot a^{-1}) \cdot x = e \cdot x = x \end{align}
(6)
\begin{align} \quad g(f(x)) = g(a \cdot x) = a^{-1} \cdot (a \cdot x) = (a^{-1} \cdot a) \cdot x = e \cdot x = x \end{align}
• Therefore $g = f^{-1}$, so $f^{-1}(x) = a^{-1} \cdot x$. $\blacksquare$