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| <math>(x, y, z) \in *_1 ~\Rightarrow~ h((x, y, z)) = (h(x), h(y), h(z)) \in *_2.\!</math>
| <math>(x, y, z) \in *_1 ~\Rightarrow~ h((x, y, z)) = (h(x), h(y), h(z)) \in *_2.\!</math>
|}
|}
A '''monoid homomorphism''' from a monoid <math>\underline{X}_1 = (X_1, *_1, e_1)\!</math> to a monoid <math>\underline{X}_2 = (X_2, *_2, e_2)\!</math> is a mapping between the underlying sets, <math>h : X_1 \to X_2,\!</math> that preserves the structure appropriate to monoids, namely, the LOCs and the identity elements. This means that the map <math>h\!</math> is a semigroup homomorphism from <math>\underline{X}_1\!</math> to <math>\underline{X}_2,\!</math> where these are considered as semigroups, but with the extra condition that <math>h\!</math> takes <math>e_1\!</math> to <math>e_2.\!</math>
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A "group homomorphism" from a group X1 = <X1, *1, e1> to a group X2 = <X2, *2, e2> is a mapping between the underlying sets, h : X1 >X2, that preserves the structure appropriate to groups, namely, the LOCs, the identity elements, and the inverse elements. This means that the map h is a monoid homomorphism from X1 to X2, where these are viewed as monoids, with the extra condition that h(x 1) = h(x) 1 for all x C X1. As it happens, the inverse elements are automatically preserved if the LOCs and the identity elements are, so a monoid homomorphism suffices to constitute a group homomorphism for a monoid that is also a group. To see why this is so, consider the following chain of equalities:
A "group homomorphism" from a group X1 = <X1, *1, e1> to a group X2 = <X2, *2, e2> is a mapping between the underlying sets, h : X1 >X2, that preserves the structure appropriate to groups, namely, the LOCs, the identity elements, and the inverse elements. This means that the map h is a monoid homomorphism from X1 to X2, where these are viewed as monoids, with the extra condition that h(x 1) = h(x) 1 for all x C X1. As it happens, the inverse elements are automatically preserved if the LOCs and the identity elements are, so a monoid homomorphism suffices to constitute a group homomorphism for a monoid that is also a group. To see why this is so, consider the following chain of equalities:
