Why the Map f(x) = ax Is Bijective in a Group

Let $G$ be a group, and let $a\in G$ be fixed. Define

$$f:G\to G,\qquad f(x)=ax.$$

The claim is that $f$ is bijective. This is the left-translation map by $a$. The key fact is that every group element has an inverse, so multiplication by $a$ can be undone by multiplication by $a^{-1}$.

The inverse map

Define another function $g:G\to G$ by

$$g(y)=a^{-1}y.$$

To check that $g$ is the inverse of $f$, compute both compositions. For every $x\in G$,

$$\begin{aligned} (g\circ f)(x)&=g(ax)=a^{-1}(ax)\\ &=(a^{-1}a)x\\ &=ex=x, \end{aligned}$$

where $e$ is the identity element of $G$. Similarly, for every $y\in G$,

$$\begin{aligned} (f\circ g)(y)&=f(a^{-1}y)=a(a^{-1}y)\\ &=(aa^{-1})y\\ &=ey=y. \end{aligned}$$

Thus $g\circ f$ and $f\circ g$ are identity functions. Therefore,

$$f^{-1}(y)=a^{-1}y,$$

so $f$ is bijective.

Direct proof: injectivity and surjectivity

A function is bijective when it is both injective and surjective.

Injectivity

Suppose $x_1,x_2\in G$ and

$$f(x_1)=f(x_2).$$

Using the definition of $f$, this says

$$ax_1=ax_2.$$

Multiply both sides on the left by $a^{-1}$:

$$\begin{aligned} a^{-1}(ax_1)&=a^{-1}(ax_2),\\ (a^{-1}a)x_1&=(a^{-1}a)x_2,\\ x_1&=x_2. \end{aligned}$$

Therefore $f$ is injective. This is exactly the cancellation law in a group.

Surjectivity

Take any $y\in G$. We need to find an $x\in G$ such that $f(x)=y$.

Choose

$$x=a^{-1}y.$$

This element belongs to $G$ because $G$ is closed under multiplication and contains $a^{-1}$. Then

$$f(x)=f(a^{-1}y)=a(a^{-1}y)=(aa^{-1})y=ey=y.$$

Since every $y\in G$ has a preimage, $f$ is surjective. Combining injectivity and surjectivity, $f$ is bijective.

A concrete example

Consider the additive group $(\mathbb{Z},+)$ and fix $a=5$. In additive notation, the corresponding function is

$$f(x)=5+x.$$

The inverse operation is addition by $-5$:

$$f^{-1}(y)=y-5.$$

For example,

$$f(7)=5+7=12,$$

and applying the inverse gives

$$f^{-1}(12)=12-5=7.$$

Every integer occurs exactly once as an output, so this translation of the integers is bijective. In multiplicative notation, the same argument uses $a^{-1}$ instead of $-a$.

As a finite noncommutative example, take the group $S_3$ of permutations of three objects and fix a permutation $a$. The map $x\mapsto ax$ rearranges the six elements of $S_3$. Its inverse is $y\mapsto a^{-1}y$, regardless of whether $S_3$ is commutative.

Left and right multiplication

The order of multiplication matters in a nonabelian group. The map in the question is left multiplication:

$$L_a(x)=ax.$$

It is bijective, with inverse $L_{a^{-1}}$.

There is also a right-translation map,

$$R_a(x)=xa,$$

which is likewise bijective. Its inverse is

$$R_{a^{-1}}(y)=ya^{-1}.$$

For right multiplication, the cancellation proof must multiply on the right. For example, from $x_1a=x_2a$ one obtains $x_1=x_2$ by multiplying by $a^{-1}$ on the right.

Common mistakes

Treating $a$ as a variable

The statement assumes that $a$ is fixed. If $a$ changes with $x$, the expression may define a different function, and bijectivity is no longer automatic.

Forgetting the codomain

The usual statement is $f:G\to G$. If the codomain is changed, the same formula might not be surjective onto that codomain. For example, a map from $G$ into a larger set containing $G$ need not be surjective onto the larger set.

Reordering factors

In a general group, $ab$ need not equal $ba$. The correct inverse of $x\mapsto ax$ is $y\mapsto a^{-1}y$, not necessarily $y\mapsto ya^{-1}$. The latter is still a valid expression in the group, but it is not generally the inverse of left multiplication.

Using only cancellation

Cancellation proves injectivity, but not by itself surjectivity. To prove surjectivity, explicitly solve $ax=y$, obtaining $x=a^{-1}y$.