Trivial Group is Initial Object

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Let $\mathbf{Grp}$ be the category of groups.

Let $1 = \set e$ be the trivial group.

Then $1$ is an initial object of $\mathbf{Grp}$.


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

By Group Homomorphism Preserves Identity, any hypothetical group homomorphism $\phi: 1 \to G$ must satisfy:

$\map \phi e = e_G$

Let us define the mapping $\phi$ in this way.

By Equality of Mappings, only one such mapping $1 \to G$ can exist, establishing uniqueness.

Now to verify that $\phi$ is actually a group homomorphism.

Since $1$ has only the element $e$, this is verified by:

\(\ds \map \phi e \circ \map \phi e\) \(=\) \(\ds e_G \circ e_G\) Definition of $\phi$
\(\ds \) \(=\) \(\ds e_G\) Definition of Identity Element
\(\ds \) \(=\) \(\ds \map \phi e\)
\(\ds \) \(=\) \(\ds \map \phi {e * e}\)

where we used $*$ to denote the group operation on $1$.

Thus $\phi$ is a group homomorphism.

We have thus established that there is a unique morphism $1 \to G$ in $\mathbf{Grp}$ for all $G$.

That is, $1$ is an initial object in $\mathbf{Grp}$.