If $A$ is a normal subgroup of $G$ and $B$ is a subgroup of $G$ , then $A\cap B$ is a normal subgroup of $B$ Suppose $A$ is a normal subgroup of $G$ and $B$ is a subgroup of $G$. Show that $A\cap B$ is a normal subgroup of $B$.
 A: $B \cap A$ is the kernel of $B \to G \to G/A$.
A: Let me expand on Martin's answer, since you may not have encountered the necessary results to make use of it, yet. To understand these results, you'll need to be somewhat familiar with group homomorphisms, the kernels of such homomorphisms, and some of the equivalent definitions of a normal subgroup.

Proposition: Given a group $G$ and a normal subgroup $N$ of $G$, denote by $G/N$ the set of left cosets of $N$ in $G$--that is the set of $aN:=\{an:n\in N\}$ with $a\in G$. The following hold for all $a,b\in G$:

(i) $aN=N$ if and only if $a\in N.$
(ii) $aN=bN$ if and only if $b^{-1}a\in N.$
(iii) $aNbN:=\{an_1bn_2:n_1,n_2\in N\}=abN.$
(iv) $G/N$ is a group under the operation defined in (iii), with identity $N$ and with $(aN)^{-1}=a^{-1}N$.

Proof of Proposition:
(i) If $a\in N$, then $an\in N$ for all $n\in N$, so $aN\subseteq N$. In particular, we have $a\in N$, so $a^{-1}\in N$ since $N$ is a subgroup. Given $n\in N$, we then have $a^{-1}n\in N$ by closure, so $n=aa^{-1}n\in aN$ by definition, whence $N\subseteq aN$, and so $aN=N.$
On the other hand, suppose $aN=N$. Since $N$ is a subgroup, then the identity of $G$ is in $N$, so $a$ is in $aN$ by definition, and so $a\in N$.
(ii) If $b^{-1}a\in N$, then by (i), $b^{-1}aN=N$. Thus, for each $n_1\in N$ there is some $n_2\in N$ with $b^{-1}an_1=n_2$, or equivalently with $an_1=bn_2$, whence $an_1\in bN$, and so $aN\subseteq bN.$ Since $b^{-1}a\in N$ and $N$ is a subgroup of $G$, then $a^{-1}b=(b^{-1}a)^{-1}\in N$. From this fact, similar arguments will show that $bN\subseteq aN$, so that $aN=bN.$
On the other hand, suppose $aN=bN$. Since $a\in aN$ as discussed in the proof of (i), then $a\in bN$, so by definition $a=bn$ for some $n\in N$, whence $b^{-1}a=n\in N$.
(iii) Since the identity of $G$ is in $N$, then we can readily write $abn=a1_Gbn\in aNbN,$ so $abN\subseteq aNbN$. Now, take $n_1,n_2\in N$. Since $n_1\in N$ and $b^{-1}\in G$ with $N$ a normal subgroup of $G$, then $b^{-1}n_1b=b^{-1}n_1(b^{-1})^{-1}\in N,$ whence $n_1b=bn$ for some $n\in N$. Since $nn_2\in N$ by closure, then $an_1bn_2=abnn_2\in abN.$ Thus, $aNbN\subseteq abN$, whence $aNbN=abN$.
(iv) In (iii) we showed that the operation on $G/N$ is well-defined. The rest follows fairly readily. $\Box$

Corollary: If $G$ is a group and $N$ a normal subgroup of $G$, define $\pi_N:G\to G/N$ by $\pi_N(a)=aN.$ ($\pi_N$ is often called the canonical projection $G\to G/N.$) Then $\pi_N$ is a homomorphism and $ker(\pi_N)=N$.
Proof of Corollary: By part (iii) of the Proposition, $$\pi_N(ab):=abN=aNbN=:\pi_N(a)\pi_N(b),$$ so $\pi_N$ is a homomorphism. By part (i) of the Proposition, $\pi_N(a):=aN=N$ if and only if $a\in N$. Since $N$ is the identity of $G/N$ by (iv), this means that $\ker(\pi_N)=N.$ $\Box$

Lemma 1: If $G,H$ are groups and $\phi:G\to H$ is a homomorphism, then $\ker\phi$ is a normal subgroup of $G$.
Proof of Lemma 1: We denote $G,H$ as multiplicative groups, and in particular denote the identity of $H$ by $1_H$. We show that for all $n\in\ker\phi$ and all $g\in G$, we have $gng^{-1}\in\ker\phi$. Indeed, $$\phi(gng^{-1})=\phi(g)\phi(n)\phi(g^{-1})=\phi(g)\phi(n)\phi(g)^{-1}=\phi(g)1_H\phi(g)^{-1}=\phi(g)\phi(g)^{-1}=1_H,$$ as desired. $\Box$

Lemma 2: Let $H,$ $G,$ and $K$ be groups, $\phi:H\to G$ and $\psi:G\to K$ homomorphisms. Then $\psi\circ\phi:H\to K$ is a homomorphism. Suppose in particular that $H$ is a subgroup of $G$ and that $\iota:H\to G$ is given by $\iota(x)=x$ for all $x\in H$. Then $\iota$ is a homomorphism (often called the canonical inclusion $H\to G$), and if $\psi:G\to K$ is a homomorphism, then $\ker(\psi\circ\iota)=H\cap\ker\psi.$
Proof of Lemma 2: For any $x,y\in H,$ $$(\psi\circ\phi)(xy)=\psi\bigl(\phi(xy)\bigr)=\psi\bigl(\phi(x)\phi(y)\bigr)=\psi\bigl(\phi(x)\bigr)\psi\bigl(\phi(y)\bigr)=(\psi\circ\phi)(x)(\psi\circ\phi)(y).$$
Now, $\iota(xy)=xy=\iota(x)\iota(y)$, so $\iota$ is a homomorphism, so $\psi\circ\iota$ is a homomorphism, and $$\begin{align}\ker(\psi\circ\iota) &= \{x\in H:(\psi\circ\iota)(x)=1_K\}\\ &= \{x\in H:\psi\bigl(\iota(x)\bigr)=1_K\}\\ &= \{x\in H:\psi(x)=1_K\}\\ &= H\cap\{x\in G:\psi(x)=1_K\}\\ &= H\cap\ker\psi,\end{align}$$ as desired. $\Box$

Now, we can put together the results above to show that if $A$ is a normal subgroup of $G$ and $B$ is a subgroup of $G$, then letting $\pi_A:G\to G/A$ be the canonical projection and $\iota:B\to G$ the canonical inclusion, we have that $A\cap B$ is the kernel of $\pi_A\circ\iota$, so is a normal subgroup.
A: For all $g \in B$, $$g (A \cap B) g^{-1} \subset gAg^{-1} \cap gBg^{-1}=A \cap B$$ since $g \in B$ implies $gBg^{-1}=B$ and $gAg^{-1}=A$ thanks to normality.
