# Intersection of two subgroups is a normal subgroup of another Intersection of two subgroups

Let $$\mathrm G$$ be a group. $$\mathrm H$$ and $$\mathrm K$$ subgroups of $$\mathrm G$$ such as $$\mathrm H\lhd\mathrm K$$. Prove that for any subgroup $$\mathrm L$$ of $$\mathrm G$$, we have

$$\mathrm H\cap\mathrm L\lhd\mathrm K\cap\mathrm L$$

I've proven that $$\mathrm H\cap\mathrm L$$ and $$\mathrm K\cap\mathrm L$$ are both subgroups of $$\mathrm G$$ what I can't prove in mathematical writing is that $$\mathrm H\cap\mathrm L$$ is a ?$$normal subgroup$$? of $$\mathrm K\cap\mathrm L$$

Is this starting a way of getting there?

$$\forall$$n$$\in$$($$\mathrm H\cap\mathrm L$$) $$\forall$$m$$\in$$($$\mathrm K\cap\mathrm L$$) : $$mnm^{-1}$$ $$\in$$ ($$\mathrm H\cap\mathrm L$$) and $$\forall$$n$$\in$$ ($$\mathrm H\cap\mathrm L$$) $$\forall$$m$$\in$$($$\mathrm K\cap\mathrm L$$) : $$mnm^{-1}$$ $$\in$$ ($$\mathrm K\cap\mathrm L$$)

• You need to show that if $m\in K\cap L$ and $n\in H\cap L$ then $mnm^{-1}\in H\cap L$. Did you prove it or not?
– Mark
Commented Oct 23, 2019 at 21:17
• I did indeed prove that, then what are you saying? Is solved just by that? Cause it's seems a little trivial then Commented Oct 23, 2019 at 21:20
• Well, yes. What is your definition of a normal subgroup? There are many equivalent definitions, one of them is that $N\trianglelefteq G$ if for all $g\in G, x\in N$ we have $gxg^{-1}\in N$. (assuming that $N$ is a subgroup of $G$ of course)
– Mark
Commented Oct 23, 2019 at 21:24
• FYI: You don't have to place a $ on both sides of every symbol; instead, it is enough to put them at both sides of a string of symbols, like $\alpha\beta\gamma$ for$\alpha\beta\gamma$. Commented Oct 23, 2019 at 21:31 • I got that. What's keep bugging my ear it's that$\mathrm H$is already a normal subgroup of$\mathrm K$and$\mathrm K$is a subgroup of$\mathrm G$. Wich I think makes$\mathrm H\subseteq\mathrm K\subset\mathrm G\$ Commented Oct 23, 2019 at 21:32

Given $$x\in K\cap L$$ then $$xLx^{-1} = L$$ because $$x\in L$$ while $$xHx^{-1}=H$$ because $$x\in K$$ and $$K$$ is contained in the normalizer of $$H$$. Therefore, because $$g\mapsto xgx^{-1}$$ defines a permutation on $$G$$, namely the $$G$$ inner automorphism induced by $$x$$, we must have
$$x(H\cap L)x^{-1}=xHx^{-1}\cap xLx^{-1}=H\cap L$$