Let $H\triangleleft G$ and $K\triangleleft G$ show that $H\cap K\triangleleft G$

Question

Please check my proof

If $H\triangleleft G$ then $gHg^{-1}=H $ for every $g\in G $

If $K\triangleleft G$ then $gKg^{-1}=K$ for every $g\in G$

Then $H\cap K = gHg^{-1}gKg^{-1}=gHKg^{-1}$

Then $H\cap K\triangleleft G$

Answer 1

The third line is wrong: you seem to be claiming that $H\cap K=HK$, which isn't true (e.g. try $G=\mathbb Z/6\mathbb Z$, $H=\langle2\rangle$, $K=\langle3\rangle$).

What you should be doing is proving that $g(H\cap K)g^{-1}=H\cap K$. Hint: Take $a\in H\cap K$; then $a\in H$ and $a\in K$, so what can you say about $gag^{-1}$?

Answer 2

You proof is flawed as $$H\cap K \subseteq H\subseteq HK$$ where equality does not need to hold so you cannot say that.

A better method is this to show that for $x\in H\cap K$ we have $gxg^{-1}\in H\cap K$ for all $g$.

This is trivial as $x\in H$ gives that $gxg^{-1}\in H$ and $x\in K$ gives that $gxg^{-1}\in K$ which means that $gxg^{-1}\in H\cap K$.