Group with Zero Element is Trivial

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Theorem

If a group $\left({G, \circ}\right)$ has a zero element, then $\left({G, \circ}\right)$ is the Trivial Group.

Let $z \in G$ be a zero element, and $e \in G$ be the identity element of $G$.

Let $x \in G$ be any arbitrary element of $\left({G, \circ}\right)$.

Then:

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle x$	$=$	$\displaystyle x \circ e$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Group axioms: G2: Identity Axiom
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle x \circ \left({z \circ z^{-1} }\right)$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Group axioms: G3: Inverse Axiom
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \left({x \circ z}\right) \circ z^{-1}$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Group axioms: G1: Associativity Axiom
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle z \circ z^{-1}$	$\displaystyle $	$\displaystyle $	$\displaystyle $	as $z$ is a zero element: $x \circ z = z$
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle e$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Group axioms: G2: Identity Axiom

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.

$\blacksquare$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x\)	\(=\)	\(\displaystyle x \circ e\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Group axioms: G2: Identity Axiom
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x \circ \left({z \circ z^{-1} }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Group axioms: G3: Inverse Axiom
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({x \circ z}\right) \circ z^{-1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Group axioms: G1: Associativity Axiom
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle z \circ z^{-1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	as $z$ is a zero element: $x \circ z = z$
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle e\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Group axioms: G2: Identity Axiom