Let $G_1$ be a simple group, that is the only normal subgroups of $G_1$ are itself and the trivial subgroup. If $\phi : G_1 \rightarrow G_2$ is a group homomorphism, does that mean $\phi$ is injective? Could someone explain?
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It does not mean that $\phi$ is injective. Since the kernel of $\phi$ must be a normal subgroup, the kernel can only be trivial or $G_1$. If the kernel is trivial, $\phi$ is injective. Otherwise, if the kernel is $G_1$, then $\phi$ is the zero map and is not injective (unless $G_1$ is, itself, trivial). – Michael Burr Aug 18 '15 at 18:39
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Looking at $\ker \phi\lhd G_1$ we note that $\phi$ is either injective or trivial.
Hagen von Eitzen
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Thank you so much Hagen. So if $kerϕ$= $G_1$, then $ϕ$ is not injective, why cannot we have this case? – Fabian Aug 18 '15 at 14:25
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We can have this case. That is the case where $\phi$ is trivial that he refers to. – Brian Moths Aug 18 '15 at 17:06
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First of all let's make an observation.
Oss: Let $f:G \rightarrow H$ an homomorphism of groups. Then $\ker f$ is a normal subgroup of $G$.
In fact if $k \in \ker f$ for all $g \in G$ we have that $f(gkg^{-1})=f(g)f(k)f(g^{-1})=f(g)f(g^{-1})=e$. Then $g \ker f \, g^{-1} \subset \ker f $ that implies $\ker f$ is normal in $G$
Now in your case $G$ is simple so $\ker f$ can only be $\{e\}$ or all $G$, so the homomorphism is injective or is the null one.
Sabino Di Trani
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