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Let $H$ be a subgroup of group $G$ and let $N$ be a normal subgroup of $G$. Choose the correct statement :

  1. $H\cap N$ is a normal subgroup of both $H$ and $N$
  2. $H\cap N$ is a normal subgroup of $H$ but not necessarily of $N$
  3. $H\cap N$ is a normal subgroup of $N$ but not necessarily of $H$
  4. $H\cap N$ need not to be a normal subgroup of either  $H$ or $N$
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$H ∩ N$ is not necessarily a normal subgroup of $N.$

Take $S_3$ for example

$S_3$ is the normal subgroup of $S_3.$

and $(e,(1,2))$ is the  subgroup of $S_3.$

So, let $N = S_3$

and $H = (e, (1,2))$

Then we find that $H∩N = H = (e,(1,2))$ which is not a normal subgroup of $N.$ (Think)

So, $H∩N$ need not necessarily be a normal subgroup of $N.$

So, either option B or option D is true.

Now, consider the theorem

$N$ is a normal subgroup of $G$ iff $gNg^{-1} ⊆ N$ for all $g∈G.$

Now, let  $x$ be an arbitrary element in $N.$

then $gxg^{-1} \in N.$

Now, $H$ is a subgroup of $G.$

So, for all $g∈ H$ and $x$ an arbitrary element in $H,$

we have $gx∈ H.$

Again from the definition of a subgroup we have

$gxg^{-1}  ∈ H.$

So, we get that for any arbitrary element $x$ in $H∩N$

$(gxg^{-1}) ∈ H∩N$  for all $g∈ H.$

Hence $(g H∩N g^{-1} ) ⊆  H∩N,$ for all $g ∈ H$

So, $H∩N$ is a normal subgroup of H.

Thus, Option B is the correct answer.
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