Prove that the subgroup $H$ of a group $G$ is a normal subgroup of the normalizer $N(H)$

abstract-algebragroup-theorynormal-subgroupssolution-verification

This question asked me to show that the normalizer $$N(H) = \{g \in G \mid gHg^{-1} = H \} \subset G$$ Is a subgroup containing $H$ and that $H$ is a normal subgroup of $N(H)$. I've shown that $N(H) < G$ and $H \subset N(H)$ I'm just having some difficulty parsing everything for this last part.

My first instinct is to say this is almost trivial since the definition of a normal subgroup $N < G$ is $\forall g \in G$ $$ \{gng^{-1} \mid n \in N \} = gNg^{-1} = N$$ And we see that the normalizer of $H$, $N(H)$, is the largest subgroup of $G$ that contains $H$ where $H$ is normal. So given that, could I just say that by definition of the normalizer we can see that $H$ fulfills the requirements for a normal subgroup?

Best Answer

  1. $N_G(H)$ is a subgroup of $G$: Clearly $1_G\in N_G(H)$. This is because $1_GH1_G^{-1}=1_GH1_G=H$. So $N_G(H)$ is a non-empty set. Then we apply the subgroup test: $N_G(H)$ is a subgroup of $G$ if $xy^{-1}\in N_G(H)$ for all $x,y\in N_G(H)$. Suppose that $x,y\in N_G(H)$. Then $(xy^{-1})H(xy^{-1})^{-1} = xy^{-1}Hyx^{-1}$. Since $y\in N_G(H)$, we have $yHy^{-1}=H$, so $H=y^{-1}Hy$ (obtained from right-multiplying by $y$ and left-multiplying by $y^{-1}$). This means that $xy^{-1}Hyx^{-1} = xHx^{-1} = H$ since $x\in N_G(H)$. Thus $xy^{-1}\in N_G(H)$ and the subgroup test tells us that $N_G(H)$ is a subgroup of $G$.
  2. $H\subseteq N_G(H)$: Let $h\in H$. Notice that $hH=H=Hh$. Thus $hHh^{-1}=H$ and $h\in N_G(H)$. Note: $gHg^{-1}=H$ if and only if $gH=Hg$, so the normaliser could have been defined as $N_G(H) := \{g\in G\ |\ gH=Hg\}$.
  3. $H$ normal subgroup of $N_G(H)$: Clearly $H$ is a subgroup of $G$ and $H$ is a subset of $N_G(H)$, so $H$ is a subgroup of $N_G(H)$. Then we apply the normal subgroup test: $H$ is a normal subgroup of $N_G(H)$ if $nhn^{-1}\in H$ for all $n\in N_G(H)$ and $h\in H$. Suppose that $n\in N_G(H)$ and $h\in H$. Then $nHn^{-1} =H$. Then it immediately follows that $nhn^{-1}\in H$ and we're done.
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