Exercise 26b.

Let $S$ be a semigroup with an element $e$ such that $ea = a = ae$ for all $a \in S$ . Show that $e$ is unique.

Proof

Let $e_1, e_2 \in S$ and let them have the identity element property, that is: $e_1a=a=ae_1$ $e_2a=a=ae_2$

Thus, by transitivity, we get $e_1a = e_2a$ . Multiplying (in the group product sense) both side by $a^{-1}$ from the right, we get

$\begin{align} (e_1a)a^{-1} = (e_2a)a^{-1} & \iff e_1(aa^{-1}) = e_2(aa^{-1}) \\ & \iff e_1e = e_2e \\ & \iff e_1 = e_2 \end{align}$

We have shown that if there are two elements with the identity element property, they have to be the same element, the proof is complete.

$\square$

Additional thoughts

Admittedly, I solved this using the cancellation law, but the book never proves the cancellation law of groups. Here is a link to the proof. When I self-study I start questioning the most basic steps, which I often find to be a bad habit. Here, I questioned why can we even multiply the equation with an element of the group. I've proven and convinced myself of this at some point in the past, so I'm too lazy to do it again. Here is an answer to this question as well--it's nice to be reminded that we are dealing with functions; however obvious it may be, I keep losing that context while in the group/algebra context. And yes this can be proven by contradiction. Every time I see an opportunity to prove by contradiction I remember Andrej Bauer (I believe) saying that we should practice logical hygiene, and I try to prove the thing directly.