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I'm trying to understand normal subgroups and kernels of homomorphisms. Normal subgroups are defined as such:

$gHg^{-1}=H~~~\forall g \in G$

While i was trying to see which subgroups are normal, to verify the above statment, i should run through all elements of $G$.

To find a shorter way, I come up with the below:

$hGh^{-1}\cap H=\emptyset~~~\forall h \in H $ (wrong)

$h(G \setminus H)h^{-1}\cap H=\emptyset~~~\forall h \in H $ (correct)

This is basically telling that, if H is normal, then $hGh^{-1}$ is in $G \setminus H$.

Assume the contrary,

$h_{0}Gh_{0}^{-1}=h_{1}$ implies $h_{0}G=h_{3}$ implies $G=h_{4}$ or $G=H$ which is a contradiction.

Could anybody prove the above statement from the first, or possible vice versa?

Second one cheaper since H has fewer elements to run through, so i would prefer to write tests over the second one.

Regards.

  • 0
    There are other ways to check whether a given subgroup $H$ is normal in a given group $G$. You can check whether the left cosets of $H$ in $G$ are also right cosets. You can try to find a homomorphism from $G$ to some other group $K$ with kernel $H$. If you already know the conjugacy classes of $G$ you can check whether $H$ is a union of conjugacy classes.2012-03-05
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    The statement $h(G \setminus H)h^{-1} \cap H = \phi$ for all $h \in H$ is true for all subgroups $H$ of $G$, not just normal subgroups.2012-03-05
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    ... and you can just check $gHg^{-1}=H$ where $g$ runs over a set of *generators* of $G$ (you don't need to check every element of $G$).2012-03-06

3 Answers 3