Math and science::Algebra::Aluffi

Groups. Monomorphism equivalents.

$monomorphism ⟺ kernel = {e} ⟺ set-injective$

Let $G$ and $H$ be groups, and $φ : G \to H$ be a group homomorphism.

The following are equivalent:

$φ$ is a monomorphism.
$\ker φ = {e_{G}}$ .
$φ$ 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) $⟹$ 2): Let $φ : G \to H$ be a group monomorphism. Being a monomorphism, any two morphisms $σ_{1} : Z \to G$ and $σ_{2} : Z \to G$ can be distinguished by their composition $φ \circ σ_{1}$ and $φ \circ σ_{2}$ . Let $\ker φ$ be our $Z$ , and consider the two group morphisms $ι : \ker φ \to G$ and $ϵ : \ker φ \to G$ (the inclusion and the trivial map respectively). Both of these maps are the same trivial map $\ker φ \to G$ , sending every element to $e_{G}$ . As they are indistinguishable, we must have $ι = ϵ$ , and so $\ker φ$ must only contain $e_{G}$ .
2) $⟹$ 3): Let $φ : G \to H$ be a group homomorphism with $\ker φ = {e_{G}}$ . Let $g_{1}, g_{2} \in G$ be two elements that map to the same element $h_{1}$ in $H$ . In other words, $φ (g_{1}) = φ (g_{2}) = h_{1}$ . 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 $φ (g_{x}) = φ (g_{1}^{- 1}) φ (g_{2}) = h_{1}^{- 1} h_{1} = e_{H}$ . And so $g_{x} \in \ker φ$ . But we know that $\ker φ$ only contains $e_{G}$ , and so $g_{1} = g_{2}$ . And so $φ$ is injective.
3) $⟹$ 1): Being injective is sufficient to be a monomorphism in caterogy $Set$ . Category $Grp$ 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 $Grp$ .

Alternative proof for 2) $⟹$ 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

ϕ

, it doesn't matter whether

g_{1}

g_{2}

to map to

h_{1}

or whether

g_{1}^{- 1}

g_{2}^{- 1}

maps to

h_{1}^{- 1}

. Thus we must have the combination:

$g_{1} g_{2}^{- 1} \in \ker φ$
$g_{1} g_{1}^{- 1} \in \ker φ$
$g_{2} g_{2}^{- 1} \in \ker φ$
$g_{2} g_{1}^{- 1} \in \ker φ$

If $\ker φ = {e_{G}}$ , 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 $\ker φ$ . This effect cascades to all other elements in $G$ , as now we have a second element in $\ker φ$ 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 $H$ .

Source

Aluffi. p84

Groups. Monomorphism equivalents.

monomorphism⟺kernel={e}⟺set-injective

Alternative proof for 2) ⟹ 3)

Relation to Lagrange's Theorem

Source

$monomorphism ⟺ kernel = {e} ⟺ set-injective$

Alternative proof for 2) $⟹$ 3)