Groups. Monomorphism equivalents.
\( \text{monomorphism} \iff \text{kernel} = \{e\} \iff \text{set-injective} \)
Let \( G \) and \( H \) be groups, and \( \varphi : G \to H \) be a group homomorphism.
The following are equivalent:
- \( \varphi \) is a monomorphism.
- \( \ker \varphi = \{ e_G\} \).
- \( \varphi \) is injective as a set-function.
The implication from 2) to 3) reminds me of Lagrange's theorem.
Can you remember the proofs?
Proof.
- 1) \( \implies \) 2)
- Let \( \varphi : G \to H \) be a group monomorphism. Being a monomorphism, any two morphisms \( \sigma_1 : Z \to G \) and \( \sigma_2 : Z \to G \) can be distinguished by their composition \( \varphi \circ \sigma_1 \) and \( \varphi \circ \sigma_2 \). Let \( \ker \varphi \) be our \( Z \), and consider the two group morphisms \( \iota : \ker \varphi \to G \) and \( \epsilon : \ker \varphi \to G \) (the inclusion and the trivial map respectively). Both of these maps are the same trivial map \( \ker \varphi \to G \), sending every element to \( e_G \). As they are indistinguishable, we must have \( \iota = \epsilon \), and so \( \ker \varphi \) must only contain \( e_G \).
- 2) \( \implies \) 3)
- Let \( \varphi : G \to H \) be a group homomorphism with \( \ker \varphi = \{ e_G \} \). Let \( g_1, g_2 \in G \) be two elements that map to the same element \( h_1 \) in \( H \). In other words, \( \varphi(g_1) = \varphi(g_2) = h_1 \). Now consider the element "from \( g_1 \) to \( g_2 \)" by which we mean \( g_1^{-1}g_2 \). We will refer to this element as \( g_x \). We have \( \varphi(g_x) = \varphi(g_1^{-1})\varphi(g_2) = h_1^{-1}h_1 = e_H \). And so \( g_x \in \ker \varphi \). But we know that \( \ker \varphi \) only contains \( e_G \), and so \( g_1 = g_2 \). And so \( \varphi \) is injective.
- 3) \( \implies \) 1)
- Being injective is sufficient to be a monomorphism in caterogy \( \cat{Set} \). Category \( \cat{Grp} \) places additional requirements on functions in order to qualify as morphisms, and so the ability of a set injective function to not obscure by left-composition is strong enough to not obscure by left-composition in \( \cat{Grp} \).
Alternative proof for 2) \( \implies \) 3)
Note that as \( g_1 \) and \( g_2 \) both map to \( h_1 \), their inverses must both map to \( h_1^{-1} \). When considering \( h_1 h_1^{-1} = e_H \) through the preimage of \( \phi \), it doesn't matter whether \( g_1 \) or \( g_2 \) to map to \( h_1 \) or whether \( g_1^{-1} \) or \( g_2^{-1} \) maps to \( h_1^{-1} \). Thus we must have the combination:- \( g_1 g_2^{-1} \in \ker \varphi \)
- \( g_1 g_1^{-1} \in \ker \varphi \)
- \( g_2 g_2^{-1} \in \ker \varphi \)
- \( g_2 g_1^{-1} \in \ker \varphi \)
If \( \ker \varphi = \{ e_G \} \), then we must have \( g_1 = g_2 \).
Relation to Lagrange's Theorem
Having two elments from \( G \) map to the same element in \( H \) forces their difference, \( g_1^{-1}g_2 \) and \( g_2^{-1}g_1 \), to be in \( \ker \varphi \). This effect cascades to all other elements in \( G \), as now we have a second element in \( \ker \varphi \) that can be applied to all other elements in \( G \). In effect, if two elements in \( G \) map to the same element in \( H \), then all elements in \( G \) must come in pairs that map to the same element in \( H \). This is the same as saying that the order of \( G \) must be a multiple of the order of \( H \).