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Let $N$ and $K$ be subgroup of a group $G$. If $N$ is normal in $G$, prove that $N \cap K$ is a normal subgroup of $K$.

Since $N$ is normal in $G$, we have $Ng = gN$ for some $g \in G$. Also observe that $N \cap K \subseteq N$ and $N \cap K \subseteq K$. I must show that $N \cap K)k = k(N \cap K)$ for some $k \in K$, to prove that $N \cap K$ is normal.

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$k(N\cap K)k^{-1}=kNk^{-1}\cap kKk^{-1}=\cdots$ – anon Feb 16 '14 at 23:54
up vote 0 down vote accepted

Let $q \in N \cap K$. You want to show that conjugating $q$ by some $k \in K$ lands you back in $N \cap K$, or in other words, that $kqk^{-1} \in N \cap K$.

Now note that since $q \in N \cap K$, we know that $q \in N$ and that $q \in K$. So thus $kqk^{-1} \in N$ since $N$ is normal in $G$, and also $kqk^{-1} \in K$, because $k$ and $q$ are in $K$. So thus $kqk^{-1} \in N \cap K$.

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A subgroup $N\subseteq G$ is normal in $G$ iff for every $g\in G$ and $n\in N$ the element $g^{-1}ng$ is in $N$. (In Fakt this means $g^{-1}Ng\subseteq N$ for every $g\in G$. Substitute $g\mapsto g^{-1}$ and you get $N\subseteq g^{-1}Ng$, therefore $Ng=gN$ for every $g\in G$)

It should be easy to prove with this equivalence, that $N\cap K$ is normal in $K$.

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