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Let $H$ and $K$ be normal subgroups of a group $G$. Prove that $H \cap K $ is a normal subgroup of $G$


Showing $H \cap K$ is a subgroup

$$a,b\in H\cap G\Rightarrow ab^{-1} \in H \wedge ab^{-1}G \Rightarrow ab^{-1}\in H \cap G$$

Having trouble with $H\cap K\unlhd G$

trying to use the following definitions

$N \unlhd G$

$$\begin{aligned} a^-1Na=N, \text{ forall } a\in G \\ aNa^{-1}=N, \text{ forall } a\in G \end{aligned} $$

since $g\in H \cap K$

$$ \begin{aligned} gHg^{-1}=H \\gKg^{-1}=K \end{aligned} $$


I did read a previous question this and could not convince myself of the proof.

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    $\begingroup$ $H\cap K\subseteq H\unlhd G$ $\endgroup$ – Cornman Sep 27 '17 at 15:37
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You know that $H\unlhd G$ and $K\unlhd G$. This means that for any $g\in G$ we have $ghg^{-1}\in H$ and $gkg^{-1}\in K$, where $h\in H$ and $k\in K$. So take any $x\in H\cap K$, since $x\in H$ we have $gxg^{-1}\in H$, by the same argument we have $gxg^{-1}\in K$, so $gxg^{-1}\in H\cap K$. Thus $(H\cap K)\unlhd G$.

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