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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?

Kevin Long
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MRI
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  • If by $1$, you mean $e'$, then what you have is the definition of the kernel. Think of what it means for a function to be injective. What happens if multiple elements go to $e'$? – Kevin Long Dec 11 '17 at 06:08
  • what you mean $e'$, $1$. All of them equal $e'=1$. You must realize when $x=0\in G$. – 1ENİGMA1 Dec 11 '17 at 06:11

2 Answers2

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Theorem: Let $\phi:G\to G'$ be a group homomorphism, then $\phi$ is $1-1$ $\iff$ $Ker\;\phi =\{e\}$

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 $1-1.$

Naive
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A homomorphism $ \phi $ is injective iff its kernel is {e}. To prove this , suppose $\phi$ is injective . Then $\phi (x) = e' =\phi (e) $ means that $x=e$. Thus $ker \phi = {e} $. Now suppose $ker \phi ={e}$. Then if $ \phi (a) =\phi (b)$ it means that $\phi (a b^-1) = e' $ ie $a b^-1 = e$ and thus a=b. Thus $\phi $is injective.

Reader Manifold
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  • oo because $\phi$(G)$\subseteq$ G' the ker $\phi$ :{g$\epsilon$ |$\phi$(g)=e'} is equivilant to $phi^{-1}$({e'}) ?? – MRI Dec 11 '17 at 06:19