Identity of Group is Unique

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Theorem

Let $\struct {G, \circ}$ be a group which has an identity element $e \in G$.

Then $e$ is unique.


Proof 1

By the definition of a group, $\struct {G, \circ}$ is also a monoid.

The result follows by applying the result Identity of Monoid is Unique.

$\blacksquare$


Proof 2

Let $e$ and $f$ both be identity elements of a group $\struct {G, \circ}$.

Then:

\(\displaystyle e\) \(=\) \(\displaystyle e \circ f\) $f$ is an identity
\(\displaystyle \) \(=\) \(\displaystyle f\) $e$ is an identity

So $e = f$ and there is only one identity after all.

$\blacksquare$


Proof 3

From Group has Latin Square Property, there exists a unique $x \in G$ such that:

$a x = b$

and there exists a unique $y \in G$ such that:

$y a = b$

Setting $b = a$, this becomes:

There exists a unique $x \in G$ such that:

$a x = a$

and there exists a unique $y \in G$ such that:

$y a = a$

These $x$ and $y$ are both $e$, by definition of identity element.

$\blacksquare$


Sources