Exercise 28b.

Let $a$ be an element of a group $G$. Show that the mapping $\lambda_a\colon G \rightarrow G$ given by $\lambda_a g = ag$ for any $g \in G$ is a one-to-one correspondence.

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Proof

One-to-one correspondence is a synonym for a bijective mapping, that is, a mapping that is both surjective (onto) and injective (one-to-one). Our goal here will be to prove that the $\lambda_a$ mapping is both surjective and injective.

Proof of $\lambda_a$ being injective

Since $a \in G$, $G$ is non-empty. Let $f,g \in G$, then we have $$\lambda_a f = af$$ $$\lambda_a g = ag$$ Let's suppose now that $\lambda_a f = \lambda_a g$. We can rewrite this as $af = ag$. Since $a, f, g \in G$, and $G$ is a group, the cancellation laws hold, and thus we have $$\lambda_a f = \lambda_a g \iff af = ag \implies f = g$$ The last implication is what we needed to show, $\lambda_a$ is indeed injective

Proof of $\lambda_a$ being surjective

We need to show that for all $c \in G$, there exists a $b \in G$, such that $ab = c$. Let $c \in G$. We can write $c$ as $c = ec = (aa^{-1})c=a(a^{-1}c)$. That is, we get that $a(a^{-1}c)= c$. Since $a^{-1} \in G$ as well, the product $a^{-1}c$ is also in $G$. We can denote it like $b = a^{-1}c$, rewriting the derived above we get $ab=c$, which is what we needed to show. The mapping $\lambda_a$ is indeed surjective.

We proved that $\lambda_a$ is both injective and surjective, therefore it is bijective, which is what we needed to show. The proof is complete.

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Additional thoughts

This seemed important to me. Not only that every element of a group can be expressed as a product of two elements of the group, but this exercise tells us that every element can be expressed as a product of two elements of the group, given that one is fixed. In other words, left multiplication by $a$ defines a bijection of the group onto itself, ensuring that every element can be obtained as a left-multiplication of some element by $a$. Compare this with the set of real numbers, which is algebraically a field—can you say that all real numbers are expressible as product of other two real numbers? What about finite, or countable fields?

Here is a table representing a field with four elements, also known as Galois field with four elements (the subset highlighted in red is also a field, known as the binary field $\text{F}$ or $\text{GF(2)}$).

image from wikipedia

Seems like we have this nice property for finite fields as well (or just this particular finite field, or a "class" of finite fields of which this one is a representative?). At what point do we lose it ? Is it with completeness?