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 Question from nokks, a student: let (G,*) be a group and prove that the intersection of normal subgroups of G is itself a normal subgroup of G

Hi,

Suppose P and Q are both normal subgroups of G and let $R = P \cap Q.$ Let $x \epsilon G$ then you need to show that $R = x^{-1}Rx.$

Let $y \; \epsilon x^{-1}Rx$ they $y = x^{-1}rx$ for some $r \epsilon R.$ But since $R = P \cap Q,$ $r$ is an element of $P$ and $Q$ and hence, since $P$ and $Q$ are normal subgroups, $x^{-1}rx \epsilon P \cap Q = R.$ Thus $y \epsilon R$ and hence $x^{-1}Rx \subset R.$

Now let $s \epsilon R, x \epsilon G$ and show that $s \epsilon x^{-1}Rx.$

Penny

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