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.

  • 1
    $\begingroup$ $H\cap K\subseteq H\unlhd G$ $\endgroup$
    – Cornman
    Sep 27, 2017 at 15:37

1 Answer 1


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 $(H\cap K)\unlhd G$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.