Proof of a theorem about homomorphisms

Question

I'm having trouble finding a proof of the following theorem:

If $\phi : G \rightarrow H$ is a homomorphism, $\phi$ injective iff $ker(\phi) = ${$e_G$}.

(A link would be as appreciated as an answer, I've looked but can't seem to find a proof of it)

Answer 1

Suppose $\phi$ is injective, then $\phi(g)=e_H$ implies that $g=e_G$ since $\phi$ is injective and $\phi(e_G)=e_H$. Hence $\ker(\phi)$ is trivial. Conversely, suppose that the kernel is trivial, then $\phi(g)=\phi(h)$ implies that $\phi(gh^{-1})=e_H$, hence $gh^{-1}\in \ker(\phi)$. Since this kernel is trivial, it follows that $gh^{-1}=e_G$, or equivalently $g=h$, hence $\phi$ is injective.

Answer 2

Think of it in terms of what the kernel represents. In a sense, the kernel of a homomorphism $\phi: G \to H$ represents the "degree of failure" of injectivity of the map.

If the kernel is larger than trivial, then this means that multiple elements in $G$ get compressed to one element in $H$. For this to not happen, the kernel must be trivial.

This would make one believe that if $\phi$ has a trivial kernel, then it must be an isomorphism. However, this is not so, since we can embed groups into one another.

For example,

$$ \psi: 2\mathbb{Z} \to \mathbb{Z} $$ Is an example of a morphism that embeds one group in another, while being injective. This has the trivial kernel $ker(\psi) = \{ 0 \}$.

Proof

Consider $$\phi: G \to H$$

By the first isomorphism theorem

$$ Im(\phi) \simeq G/ker(\phi) $$

Where $Im(\phi)$ is the image of $G$ in $\phi$.

However, if $ker(\phi) = \{e\}$, then we know that

$$ G/ker(\phi) \simeq G $$

And hence,

$$ G \simeq G/ker(\phi) \simeq Im(\phi) $$

If the kernel is trivial.

Since $G$ and $Im(\phi)$ are isomorphic, we get a one-to-one map between the two, thereby proving injectivity due to a trivial kernel.