TITLE: Why is the kernel of an isomorphism always equal to the identity?
QUESTION [5 upvotes]: I'm just learning about isomorphisms. Suppose $f:G\rightarrow G'$ is an isomorphism from the groups $G$ to $G'$. Why then is the kernal of $G$ equal to $\{e_G\}$? According to my source, we have $$\text{Ker}(f)=\{g\in G:f(g)=e_{G'}\},$$ and for an isomorphism we always have $$\text{Ker}(f)=\{e_G\}.$$ I know that an isomorphism is a bijection and get the idea of a bijection, but can't yet fathom how this relates to the associated kernel.

REPLY [5 votes]: The "size" of the kernel measures the "injectivity" of a function in the following sense:
Proposition: $f$ is injective if and only if ker$(f)=\{e\}$.
If ker$(f)=\{e\}$, from the equality  $f(x)=f(y)$, you have that $f(xy^{-1})=e'$  i.e. $xy^{-1}\in$ ker$(f)$. From the hypothesis,  $xy^{-1}=e$, so you can conclude that $x=y$.
Viceversa suppose that $f$ is injective; if $x\in$ ker$(f)$, $f(x)=e'=f(e)$ and  follows   $x=e$ thanks to the injectivity .