Given $f:G\to H$ is a homomorphism then prove if the kernel $K = \{e\}$ then $f$ is injective.

Question

I don't know how to show a kernel of only the neutral element means $f$ is injective.

My ideas.

1. Using cosets show how each $Ka$ maps to a unique $f(x)$ value. Since cosets have one to one connections. And each $Ka$ has only one element.

2. Show $f$ is proving $H$ and $G$ are isomorphic, so $f$ is injective and surjective.

I don't know how to write those ideas if they are even correct.

Answer 1

Consider $g,h \in G$ such that $f(g) = f(h)$. We would like to show that $g = h$.

$$ \begin{align*} f(g) = f(h) &\implies f(g)f(h)^{-1} = e \\ &\implies f(gh^{-1}) = e \\ &\implies gh^{-1} \in \ker f = \{e\}\\ &\implies gh^{-1} = e \\ &\implies g = h \end{align*} $$ Note that the second line comes from the properties of a group homomorphism, namely that $f(g)f(h)=f(gh)$ and $f(g^{-1})=f(g)^{-1}$.

Answer 2

Suppose $f: G \to H$ is a homomorphism of groups.

If $f$ is injective the kernel is $\{e\}$: we always have $f(e)=e$ and so if $x \in \operatorname{ker}(f)$ so $f(x)=e$ we conclude that $x=e$ by injectivity.

If $\operatorname{ker}(f) = \{e\}$, assume $f(g)=f(g')$. Then

$$f(g^{-1}g')=f(g)^{-1}f(g') = f(g)^{-1}f(g)=e \text{ so } g^{-1}g' \in \operatorname{ker}(f) \text{ hence } g^{-1}g'=e$$ and multiplying both sides by $g$ from the left gives $g=g'$.