Group with Zero Element is Trivial

Theorem

Let $\struct {G, \circ}$ be a group.

Let $\struct {G, \circ}$ have a zero element.

Then $\struct {G, \circ}$ is the trivial group.

Proof

Let $e \in G$ be the identity element of $G$.

Let $z \in G$ be a zero element.

Let $x \in G$ be any arbitrary element of $\struct {G, \circ}$.

Then:

\(\ds x\)

\(=\)

\(\ds x \circ e\)

Group Axiom $\text G 2$: Existence of Identity Element

\(\ds \)

\(=\)

\(\ds x \circ \paren {z \circ z^{-1} }\)

Group Axiom $\text G 3$: Existence of Inverse Element

\(\ds \)

\(=\)

\(\ds \paren {x \circ z} \circ z^{-1}\)

Group Axiom $\text G 1$: Associativity

\(\ds \)

\(=\)

\(\ds z \circ z^{-1}\)

Definition of Zero Element: $x \circ z = z$

\(\ds \)

\(=\)

\(\ds e\)

Group Axiom $\text G 3$: Existence of Inverse Element

So whatever $x \in G$ is, it has to be the identity element of $G$.

So $G$ can contain only that one element, and is therefore the trivial group.

$\blacksquare$

Sources

• 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 4.5$. Examples of groups: Example $83$