Heyall, would appreciate some help with abstract algebra because my undergrad brain is fried from doing all the proofs my prof asked me to do. I've hit a bit of a wall with this one; it involves group homomorphisms - super basic but the proof has got to be quite sophisticated cuz my mind is blank.
there are two groups $G$ and $F$ and a mapping $\phi : G \rightarrow F$
the kernel is defined as
$ker (\phi ) := \left \{ x\in G : \phi (x) = e_{H} \right \}$ whereby $e_H$ is the identity element in $H$
to prove: $\phi$ is injective iff $ker(\phi) = \{e_{G}\}$ whereby $e_G$ is the identity element in $G$
would appreciate any tips too, thanks a bunch xx
Suppose $\phi$ is injective. As $\phi$ is a group homomorphism, $\phi(e_G) = e_H$, right? So, $g \in G$ is in the kernel of $\phi$ if and only if $\phi(g) = \phi(e_G)$, and by injectivity, $g$ is in the kernel if and only if $g = e_G$.
Assume that $\ker \phi$ consists only of $e_G$. To prove the injectivity, suppose that two elements of $G$, $g$ and $h$, satisfies that $\phi(g) = \phi(h)$ and we want to prove that $g = h$. But see this: $$e_H = \phi(g)\phi(h)^{-1} = \phi(gh^{-1})$$ and so, $gh^{-1} \in \ker \phi$, that is, $gh^{-1} = e_G$, and then, $g = h$.