Q:

Why is the kernel of an isomorphism always equal to the identity?

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.

A:

Because an isomorphism is a bijection, every element of $G'$ has exactly one preimage; in particular there is exactly one preimage of $e_{G'}$. Since $e_G$ is already a preimage of $e_{G'}$, that is the one. The kernel consists of all preimages of $e_{G'}$, so is just $\{ e_G \}$.
Or, if you prefer the reasoning in symbols, $$g \in \text{Ker}(f) \iff f(g) = e_{G'} \iff f(g) = f(e_G) \iff g = e_G.$$ The last $\iff$ uses the assumption that $f$ is a bijection (actually, just that it is injective, as that is enough for the kernel to be trivial).

A:

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 . 

