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 $(G, *)$. 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.

Theorem 1: Let $(G, *)$ 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:

\begin{align} \quad x^{-1} = y^{-1} \\ \quad x * x^{-1} = x * y^{-1} \\ \quad e = x * y^{-1} \end{align}

- Since $x * y^{-1} = e$ then by the uniqueness of the inverses with respect to $*$ we have that $x = y$, so $f$ is injective.

- Now let $y \in G$. Then $y^{-1}$ is such that:

\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. $\blacksquare$

Theorem 2: Let $(G, *)$ be a group and let $f : G \to G$ be defined for each $x \in G$ and some $a \in G$ by $f(x) = a * x$. 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:

\begin{align} \quad a * x = a * y \\ \quad a^{-1} * (a * x) = a^{-1} * ( a * y) \\ \quad (a^{-1} * a) * x = (a^{-1} * a) * y \\ \quad e * x = e * 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 $(G, *)$ is a group we must have that $a^{-1} \in G$, and so for $(a^{-1} * x) \in G$ (from the closure property of $*$ on $G$) we have that:

\begin{align} \quad f(a^{-1} * x) = a * (a^{-1} * x) = (a * a^{-1}) * x = e * x = x \end{align}

- Therefore for every $x \in G$ there exists a $y = a^{-1} * x \in G$ such that $f(y) = f(a^{-1} * x) = x$, so $f$ is surjective.

- Since $f$ is both injective and surjective then by definition $f$ is bijective. $\blacksquare$

Corollary 1: Let $(G, *)$ be a group and let $f : G \to G$ be defined for each $x \in G$ and some $a \in G$ by $f(x) = a * x$. Then $f^{-1}(x) = a^{-1} * x$. |

**Proof:**From Theorem 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 $g(x) = a^{-1} * x$ then:

\begin{align} \quad f(g(x)) = f(a^{-1} * x) = a * (a^{-1} * x) = (a * a^{-1}) * x = e * x = x \end{align}

(6)
\begin{align} \quad g(f(x)) = g(a * x) = a^{-1} * (a * x) = (a^{-1} * a) * x = e * x = x \end{align}

- Therefore $g = f^{-1}$, so $f^{-1}(x) = a^{-1} * x$. $\blacksquare$