Identity of Group is Unique

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:

\(\ds e\)

\(=\)

\(\ds e \circ f\)

$f$ is an identity

\(\ds \)

\(=\)

\(\ds f\)

$e$ 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

• 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 4.6$. Elementary theorems on groups
• 1970: B. Hartley and T.O. Hawkes: Rings, Modules and Linear Algebra ... (previous) ... (next): Chapter $1$: Rings - Definitions and Examples: $1$: The definition of a ring: Definitions $1.1 \ \text{(b)}$