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My professor proved that the kernel is a normal subgroup under a homomorphism f by saying. Let $h\in H$ by applying f to it $f(ghg^{-1})=f(g)f(h)f(g)^{-1}=e $

My question is how is this a proof that the kernel is normal. What is the reason we apply f to it ? By applying f dont we just prove that $f(ghg^{-1})$ is normal ?

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    $\begingroup$ Because by showing that, you know $ghg^{-1}$ is also in the kernel $\endgroup$ – user160738 Feb 10 '17 at 21:30
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The argument goes as:

Let $f: G \to G_1$ be the homomorphism, and let $H = \operatorname{Ker}(f) = \{g \in G \; | \; f(g)=e \}$ be its kernel.

A subgroup is normal if it is invariant by conjugation. So fix an arbitrary element $h \in H$. For all $g \in G$, is it true that $ghg^{-1} \in H$ (which would mean that $H$ is indeed invariant by conjugation)?

And then he proves it as you have done above: $$f(ghg^{-1}) = f(g)f(h)f(g^{-1}) = f(g)f(g)^{-1} = e$$

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  • $\begingroup$ This just proves that it is normal if it is a subgroup, it doesn't prove that it is a subgroup, am I right? $\endgroup$ – zabop May 24 at 16:50
  • $\begingroup$ Sure, but the OP's question was "how is this a proof that the kernel is normal". To prove that it is a subgroup just verify that it's closed w.r.t. inverse and group multiplication $\endgroup$ – AnalysisStudent0414 May 24 at 16:53
  • $\begingroup$ Oh ok, I read the title, which says subgroup, but you have a point yes. $\endgroup$ – zabop May 24 at 16:56
  • $\begingroup$ Yes, and maybe say that associativity is inherited from parent group, and identity maps to identity (so identity is part of kernel). $\endgroup$ – zabop May 24 at 16:57
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We have a group $G$ and a group homomorphism $f$ that takes elements from $G$. We also have $\ker(f) = H$. We want to show that $H$ is normal. One way of describing "$H$ is normal" is that given arbitrary $h\in H$ and $g\in G$, we must have $ghg^{-1}\in H$. By definition of $H$, $ghg^{-1}\in H$ is equivalent to $f(ghg^{-1}) = e$. So that is what we check for.

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