$G\times H$ is a group with identity $(e_G,e_H)$, coordinatewise inverses, and homomorphic coordinate projections

For $(g,h),(g^{\mathrm{\prime }},h^{\mathrm{\prime }}),(g^{\mathrm{\prime }\mathrm{\prime }},h^{\mathrm{\prime }\mathrm{\prime }})\in G\times H$, associativity in each factor gives $((g,h)(g^{\mathrm{\prime }},h^{\mathrm{\prime }}))(g^{\mathrm{\prime }\mathrm{\prime }},h^{\mathrm{\prime }\mathrm{\prime }})=(g{g}^{\mathrm{\prime }}{g}^{\mathrm{\prime }\mathrm{\prime }},h{h}^{\mathrm{\prime }}{h}^{\mathrm{\prime }\mathrm{\prime }})=(g,h)((g^{\mathrm{\prime }},h^{\mathrm{\prime }})(g^{\mathrm{\prime }\mathrm{\prime }},h^{\mathrm{\prime }\mathrm{\prime }}))$; thus the componentwise operation is associative.

For every $(g,h)\in G\times H$, $(e_G,e_H)(g,h)=(g,h)=(g,h)(e_G,e_H)$; thus $(e_G,e_H)$ is a two-sided identity.

For every $(g,h)\in G\times H$, $(g,h)(g^{-1},h^{-1})=(e_G,e_H)=(g^{-1},h^{-1})(g,h)$; so $(g^{-1},h^{-1})$ is its inverse.

Steps 1.1–1.3 verify the group axioms for $G\times H$.

For pairs $(g,h),(g^{\mathrm{\prime }},h^{\mathrm{\prime }})$, ${\pi }_G((g,h)(g^{\mathrm{\prime }},h^{\mathrm{\prime }}))=g{g}^{\mathrm{\prime }}={\pi }_G(g,h){\pi }_G(g^{\mathrm{\prime }},h^{\mathrm{\prime }})$; the same coordinatewise calculation holds for ${\pi }_H$, so both projections are homomorphisms from the group in step 2.1.

The stated identity, inverse formula, and coordinate homomorphisms follow.