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I want to prove that the unit element in a group is unique.

My attempt:

Suppose $e,e'$ are unit elements of a group $G$. By definition, we have:

$\forall g \in G: eg = g = ge$ and $e'g = g = ge'$

Thus we find that $eg = e'g$

Because $*: S \times S \rightarrow S$ (this is the multiplication on the group, but we never write it down) is a function, we can multiply both sides with an inverse of g (this inverse exists by the definition of a group and turns out to be unique). Denote this inverse with $h$. Hence, we obtain:

$(eg)h = (e'g)h \Rightarrow e(gh) = e'(gh) \Rightarrow e = e'$, using associativity and the definition of inverse elements and unit elements.

Can someone verify whether this proof looks correct? It bothers me that I did not use all the information from the definition of unit element.

Thanks in advance.

2 Answers 2

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Your proof works, but it's made slightly complicate by introducing the arbitrary element $g$ and its inverse.

A much easier way to prove it: Take a look at the element $$ee'.$$

Then, because $eg=g$ for each $g$, what happens if $g=e'$? Well, we get that $ee'=e'$ from this rule. But we also know that $ge'=g$ for each $g$, so...

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    Great answer! This proof seems to work in monoids as well, where as my proof fails because I introduce inverse elements. Can you confirm whether this is true? – 2017-02-17
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    @Math_QED Indeed. Your proof would need to be modified for monoids, but your proof actually works as long as the set contains at least **one** invertible element, and monoids do contain that, since the unit is invertible. However, the modification would then more or less transform your proof into one that looks almost exactly like mine. – 2017-02-17
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Another easy approach, first using:

Lemma: If $\;G\;$ is a group (or even a monoid) and $\;x\in G\;$ fulfills $\;x\cdot x=x\;$ , then $\;x=e\;$.

And now observe that if $\;e'\;$ is another unit of a group $\;G\;$ , then $\;e'e'=e'\;$ (as $\;e'\;$ fixes itself too) and we're done

If you want to see the proof of the above lemma put your pointer on the hidden area:

Proof: Suppose $\;e\;$ is the group's unit, then$$x=x\cdot e=x(x\cdot x^{-1})=(x\cdot x)x^{-1}=x\cdot x^{-1}=e$$