# Prove that identity element is unique

During an exam I tried to prove that the identity element of group (G.•) is unique. I approached this way:

Suppose there are two identity elements $e_1$ and $e_2$. Then:

$a^{-1}•a=e_1$

$a^{-1}•a=e_2$

$e_1=e_2$

What are the potential flaws in my proof?

• The usual definition of a two-sided identity $e$ is that $ae=ea=a$ for each $a\in G$
– MPW
Commented May 13, 2016 at 5:38
• One clear problem is that you have defined a unique inverse prior to defining a unique identity. Commented Sep 12, 2017 at 22:23

As noted by MPW, the identity element $e \in G$ is defined such that $ae=a \quad \forall a\in G$.

While the inverse does exist in the group and multiplication by the inverse element gives us the identity element, it seems that there is more to explain in your statement, which assumes that the identity element is unique.

A more standard way to show this is suppose that $e, f$ are both the identity elements of a group $G$.

Then, $e=e \circ f$ since $f$ is the identity element.

$\quad\quad\quad= f\quad$ since $e$ is the identity element.

This shows that the identity element is indeed unique.

The definition of a group does not require that $$a^{-1}\cdot a = e$$ for every identity element $$e$$. It only requires that there be at least one identity for which that is the case.

It says:

1. There is an identity element $$e$$ for which $$e\cdot a = a\cdot e = a$$.
2. Every element $$a$$ has an inverse $$a^{-1}$$ for which $$a^{-1}\cdot a = e$$.

This does not rule out the possibility that there is another identity element $$e_2$$, which has the identity property $$e_2\cdot a = a\cdot e_2 = a$$, but for which $$a^{-1}\cdot a \ne e_2$$.

We could have $$a^{-1}\cdot a = e \ne e_2$$ and the axioms would still be satisfied.

For this reason, your proof is not correct.

• This is using associative?
– MMM
Commented Dec 8, 2017 at 18:28
• I did not using associative.
– MJD
Commented Dec 8, 2017 at 19:46
• Please state some source to learn that such identity as $e_2$ is permitted. Commented Oct 3, 2020 at 19:16
• The definition of “group” on page 16 of Abstract Algebra by Dummitt and Foote (2004) says “(ii) there exists an element $e$ in $G$, called an identity of $G$, such that for all $a\in G$ we have $a\star e=e\star a=a$.” Note that they say there exists “an element” not “a unique element”, and that say that $e$ is called “an identity”, not “the identity”. The axioms do not require that there is only one identity, only that there is at least one. On page 18, the book proves (proposition 1(1)) that a group has exactly one identity. Most other elementary books on group theory will do the same.
– MJD
Commented Oct 3, 2020 at 23:50

If a set $S$ with a binary operation $\cdot$ (that is, $S$ is a magma) has a left identity $e_L$ and a right identity $e_R$, then $e_L=e_R$. This is because $e_R=e_L\cdot e_R=e_L$. This also shows that, if $S$ has an identity (i.e., a two-sided identity), then this identity is unique.

It is not true, however, that any two left identities must be identical. If we let $S:=\mathbb{N}$ and $a\cdot b:=b$ for all $a,b\in S$, then every element of $S$ is a left identity. The same goes for right identities. Only when there are both a left identity and a right identity does it imply that $S$ has a unique identity.

One of the weakest definitions of groups I have seen is the following. For a set $G$ with a binary operation $\cdot$, $(G,\cdot)$ is a group if $G$ is nonempty, $\cdot$ is associative, and for every $a,b\in G$, there exists a unique $x\in G$ such that $a\cdot x=b$. This definition gives a unique element-wise left identity $e_g$ for each $g\in G$ (namely, $e_g\cdot g=g$ for all $g\in G$). However, it turns out that $e_g=e_h$ for any $g,h\in G$, and that, if $e$ is this common left identity $e_g$ for $g\in G$, then $e$ is also a right identity, whence the unique identity of $G$.

• This is interesting. I tried proving that the weak definition indeed defines a group, but I was not able to succeed. Do you have a reference for it? Commented Nov 22, 2023 at 13:40

What happens when $e_1 • e_2$?