Group with Zero Element is Trivial

From ProofWiki
Jump to navigation Jump to search

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:

\(\displaystyle x\) \(=\) \(\displaystyle x \circ e\) Group Axiom $G \, 2$: Identity
\(\displaystyle \) \(=\) \(\displaystyle x \circ \paren {z \circ z^{-1} }\) Group Axiom $G \, 3$: Inverses
\(\displaystyle \) \(=\) \(\displaystyle \paren {x \circ z} \circ z^{-1}\) Group Axiom $G \, 1$: Associativity
\(\displaystyle \) \(=\) \(\displaystyle z \circ z^{-1}\) Definition of Zero Element: $x \circ z = z$
\(\displaystyle \) \(=\) \(\displaystyle e\) Group Axiom $G \, 3$: Inverses

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