Will the intersection of a subgroup and a normal subgroup create a normal subgroup for the main group

Question

If $G$ is a group and $H \leqslant G$ and $K \triangleleft G$, is $H \cap K \triangleleft G$? I think the intersection of $H$ and $K$ is a subgroup, and if $K \triangleleft G$ is $H\cap K \triangleleft G$ but I couldn't prove it. This is all I have done so far:

$\forall x \in H \cap K \land \forall g \in G \Rightarrow gxg^{-1} \in H \cap K$ ?

$x \in H \cap K \Rightarrow x \in H \land x\in K$

$x \in K \land g \in G \Rightarrow gxg^{-1} \in K$

but i couldn't show it

$x \in H \land g \in G \Rightarrow gxg^{-1} \in H$

Give me a hint please

Answer 1

No, $H\cap K$ might be not normal in $G$. For example take $G=D_4$, the dihedral group with $8$ elements. Letting $\rho$ be a rotation by angle $\frac{2\pi}{4}$ and $\epsilon$ any reflection we can take $K=\langle \rho^2,\epsilon\rangle\trianglelefteq G$ and $H=\langle \epsilon\rangle\leq G$. Then $H\cap K=\langle\epsilon\rangle$, and it is not normal in $D_4$.

That being said, $H\cap K$ is normal in $H$. Indeed, letting $h\in H, x\in H\cap K$ we have $hxh^{-1}\in H$ because it is a product of elements in $H$, and we have $hxh^{-1}\in K$ because $K$ is normal in $G$. So indeed $hxh^{-1}\in H$.

Answer 2

For every $a \in G$, we get:

\begin{alignat}{1} a(H\cap K) &= \{ag\mid g\in H\cap K\} \\ &= \{ag\mid g\in H\wedge g\in K\} \\ &= \{ag\mid g\in H\}\cap \{ag\mid g\in K\} \\ &= aH\cap aK \\ \tag 1 \end{alignat}

and likewise for the right cosets. Therefore, if $K\unlhd G$:

\begin{alignat}{1} a(H\cap K)\cap (H\cap K)a &= (aH\cap aK)\cap(H a\cap aK) \\ &= (aH\cap Ha)\cap aK \\ \tag 2 \end{alignat}

Now, $H \ntrianglelefteq G \Rightarrow \exists a'\in G\mid a'H\ne Ha' \Rightarrow a'H\cap Ha'\subsetneq a'H$, and thence, by $(2)$:

\begin{alignat}{1} a'(H\cap K)\cap (H\cap K)a' &= (a'H\cap Ha')\cap a'K \subsetneq a'H\cap a'K=a'(H\cap K)\\ \tag 3 \end{alignat}

whence:

$$(H\cap K)a' \ne a'(H\cap K)$$

and finally $H\cap K \ntrianglelefteq G$.

Answer 3

If you take $H\leq K$ then $H\cap K=H$. So it is enough to find a chain of subgroups $H\leq K\lhd G$ such that $H\not\lhd G$.

For example, take $G=S_3\times L$ where $L$ is any non-trivial group. Then the chain $$\{id, (12)\}\leq S_3\lhd G$$ works. That is, take $K:=S_3\lhd G$ and $H:=\{id, (12)\}\leq S_3$, and the chain works as $\{id, (12)\}\not\lhd S_3$.

(I used $S_3$ because it is the smallest group containing a subgroup which is not normal. Many, many other groups work though!)