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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$) : $mn$$m^{-1}$ $\in$ ($\mathrm K$$\cap$$\mathrm L$)

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  • $\begingroup$ 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? $\endgroup$
    – Mark
    Commented Oct 23, 2019 at 21:17
  • $\begingroup$ I did indeed prove that, then what are you saying? Is solved just by that? Cause it's seems a little trivial then $\endgroup$ Commented Oct 23, 2019 at 21:20
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    $\begingroup$ 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) $\endgroup$
    – Mark
    Commented Oct 23, 2019 at 21:24
  • $\begingroup$ 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 $\alpha\beta\gamma$. $\endgroup$
    – Shaun
    Commented Oct 23, 2019 at 21:31
  • $\begingroup$ 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 $\mathrm H$ $\subseteq$ $\mathrm K$ $\subset$ $\mathrm G$ $\endgroup$ Commented Oct 23, 2019 at 21:32

1 Answer 1

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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

$$x(H\cap L)x^{-1}=xHx^{-1}\cap xLx^{-1}=H\cap L$$

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