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Let $H$ be a subgroup of $G$, then $G$ acts on the left cosets $Cos_H$ of $H$ by left multiplication. Show that $ker(\phi)$ of the homomorphism $\phi : G \to Sym(Cos_H)$ is the largest subgroup in $H$ that is normal in $G$.

Can anyone help me with this one? Is $Cos_H$ the set of all left cosets? So $\phi$ must be a function that takes an element in $G$ and throws it on a left coset? and is $Sym(Cos_H)$ the set of al permutations with elements in $Cos_H$?

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  • $\begingroup$ Yes, it is. No, $\phi\;$ is a group homomorphism defined by the action of $\;G\;$ on what you denoted $\;Cos_H\;$ . Yes, though it'd be more accurate to describe $\;Sym(Cos_H)\;$ as the symmetric group on $\;Cos_H\;$ , meaning: all the bijectives function on this set. First get all this straight, then check again and if you're still stuck ask back. $\endgroup$ – DonAntonio Apr 27 '14 at 20:52
  • $\begingroup$ Well then I have understood it, do you have any tips because iam really stuck. @DonAntonio $\endgroup$ – user117449 Apr 27 '14 at 20:56
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First, we have a general set up: a group action $\;G\times Cos_H\to Cos_H\;$ defined by

$$g(xH):=(gx)H\;,\;\;\forall\,g,x\in G$$

As any other action, this ones determines a homomorphism $\;\phi:G\to Sym(Cos_H)\;$ , and this homomorphism's kernel is precisely

$$\ker\phi=\bigcap_{g\in G}H^g=:\text{ the core of the subgroup}\;\;H\;,\;\;H^g:=g^{-1}Hg$$

Now prove that:

(1) The above indeed is true (i.e., the kernel indeed equals that intersection of subgroups), and

(2) That kernel is the biggest subgroup contained in $\;H\;$ that is also normal in $\;G\;$ .

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  • $\begingroup$ Oh, nobody has usually worked with the core of a subgroup before this subject. You have to prove this is the maximal normal subgroup in the whole group that is contained in $\;H\;$ ...and I can't see an alternative way to prove this, though perhaps there is one. $\endgroup$ – DonAntonio Apr 27 '14 at 21:12
  • $\begingroup$ Ok I will give it a try, thanks. $\endgroup$ – user117449 Apr 27 '14 at 21:18
  • $\begingroup$ @Sodan: $$x\in\ker\phi\implies xgH=gH\;\;,\;\;\forall\,g\in G\;,\;\text{in particular we get for}\;g=1\;,\;xH=H\implies x\in H$$ $\endgroup$ – DonAntonio Apr 28 '14 at 3:07
  • $\begingroup$ is this correct? Consider the action of $G$ on the cosets of $H$ given by multiplication. This is a homomorphism $\phi$ from $G$ to the symmetric group on the set $G/H = \{ gH : g \in G \} = \{ \{ gh: h \in H \} : g \in G \}$. As such, it has a kernel $K$, those $k$ such that $kgH = gH$ for all $g \in G$. This is precisely the $k$ such that $kg = gh_g$ for some $h_g \in G$ dependent on $g$, that is $k = gh_g g^{-1} \in gHg^{-1}$. In other words, $\bigcap_{g\in G} gHg^{-1} = \ker(\phi)$. $\endgroup$ – user117449 Apr 28 '14 at 20:21
  • $\begingroup$ Would you consider looking at my answer above please? $\endgroup$ – user117449 Apr 29 '14 at 14:51

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