Lemma 1: Let $G$ be a group and let $H,K,N \leq G$ with $N$ normal in $H$. Then $N \cap K$ is normal in $H \cap K$. Proof: Let $a \in H \cap K$. Then $a(N \cap K) = aN \cap aK = Na \cap Ka = (N \cap K)a $because $a \in H$ and $a \in K$.
Throughtout the link they use this method :$a(N \cap K) = aN \cap aK = Na \cap Ka = (N \cap K)a$ to show that a certain subgroup is a subgroup of another subgroup.
I never seen this before.
In this proof I would have done something like this:
$N\cap K$ is normal since the intersection of a normal subgroup and subgroup is normal.
Let $a\in N\cap K$ then $a\in N $and $ a\in K$. $N\leq H$ so $a\in H$ too. So we have $a\in H\cap K$. Thus $N \cap K \trianglelefteq H \cap K$.
Can anyone explain why showing that $a\in H\cap K$ commutes with $N\cap K$
$$a(N \cap K) = aN \cap aK = Na \cap Ka = (N \cap K)a$$
implies $N\cap K \leq H\cap K$ .
Thanks :D I appreciate it
Let $x\in N\cap K$. Then $a(N\cap K) = (N\cap K)a$ means that $a(N\cap K)a^{-1} = N\cap K$, i.e., that $axa^{-1} \in N\cap K$, the exact condition for normality.