If $\phi$ is a homomorphism from the group $G$ with identity $e$ to the group $G'$ with identity $e'$, then $\phi$ is injective if and only if $\ker\phi=$?
I think the $Ker\phi = \{x\in G:\phi(x)=1\}$
because you are looking at an identity then it is all reals $0$ does not work a because if $x=0, e'x=0$
but then i realize it is injective how does that affect the kernel?
Best Answer
Proof:
Firstly assume $\phi$ be $1-1$.
Let $x\in Ker\;\phi$, then $\phi(x)=e'=\phi(e)\implies x=e$.
Hence, $Ker\;\phi=\{e\}$
Conversely, assume $Ker\;\phi=\{e\}$.
Let, $\phi(x)=\phi(y)\\\implies \phi(x)-\phi(y)=e'\\\implies\phi(x-y)=e'\\\implies x-y=e'\\\implies x=y$
Thus, $\phi$ is $1-1.$