$H,K$ are normal in $G$, then $HK$ is normal in $G$ (product of normal subgroups is normal) This is a proof I couldn't find anywhere. Could somebody give me a help?
I need this to show that $$\frac{H}{H\cap K}\cong \frac{HK}{K}$$
but to form the quotient group I need first to show that $H\cap K$ is normal in $H$ and $K$ is normal in $HK$.
 A: This is not hard to see :
$$
gHKg^{-1} = (gHg^{-1})(gKg^{-1}) \subseteq HK. 
$$
That is, any element of the form $ghkg^{-1}$ can be written in the form $(ghg^{-1})(gkg^{-1})$, the first factor being in $H$ and the second in $K$. 
Hope that helps,
A: We have $g(hk)g^{-1}=(ghg^{-1})(gkg^{-1})\in HK$
A: I think you posed three different questions in one. 
Let $H$ and $K$ be normal subgroups in $G$. Then: 


*

*$HK$ is normal in $G$;

*$H\cap K$ is normal $H$; 

*$K$ is normal in $HK$.
Answers: For answers 1 and 3 below, I will assume you already know how to prove that $HK$ is a subgroup of $G$. It is easy to prove using the fact that $H$ (or $K$) is a normal subgroup of $G$.


*

*As already mentioned in Patrick's and Guerlando's answer.  Given any $g\in G$,  $gHKg^{-1}= (gHg^{-1})(gKg^{-1}) \subseteq HK$.  So $HK$ is normal in $G$. 

*Since $H$ and $K$ are normal in $G$, we know that given any $g\in G$, 
$gHg^{-1} \subseteq H$ and $gKg^{-1} \subseteq K$. So we have 
$g(H\cap K)g^{-1} = gHg^{-1} \cap gKg^{-1} \subseteq (H\cap K)$. 
So $(H\cap K)$ is normal in $G$. Since $(H\cap K) \subseteq H \subseteq G$, we have that $(H\cap K)$ is normal in $H$. 

*Since $K$ is normal in $G$ and $K \subseteq HK \subseteq G$, we have that $K$ is normal in $HK$. 
A: Hint: Actually you only need to show that $HK$ is a subgroup of $G$ and that $HK= KH$. 
Claim 1: $HK$ is a subgroup of $G$ if, and only if, $HK=KH$. 
Proof: See this answer.
Corollary: If $H$ or $K$ is normal in $G$ then $HK$ is a subgroup.
Proof: Say $H \triangleleft G$ and $K$ is any subgroup of $G$. Just show that $Hk=KH$. Let $\alpha = hk \in HK$. Then $$\alpha = hk = kk^{-1}hk = k \beta$$ 
where $\beta = k^{-1}h k \in H$. Thus $\alpha = k\beta \in KH$, and it follows that $HK \subseteq KH$. Try and show the other inclusion. 
To attack the question in hand, notice that $H \triangleleft G \implies H \triangleleft KH$ and consider the homomorphism $\varphi : KH \to KH/H$ and let $\varphi|_K$ the restriction to $K < KH$. That is, $$\begin{align} \varphi|_K: K &\to \frac{KH}{H}\\k&\mapsto kH \end{align}$$
show that $\ker \varphi|_K = H \cap K$ and that $\varphi|_K$ is surjective.
A: As Aaron Maroja indicates, it is sufficient to show $H \cap K$ is the kernel of a homomorphism (which then forces it to be normal), but it is easy to show this directly:
Suppose $K \lhd G$, and $H$ is any subgroup of $G$. Then $H \cap K \lhd H$.
To see this, let $h \in H$ be arbitrary. We consider $h(H \cap K)h^{-1} = \{hxh^{-1}: x \in H \cap K\}$.
Since $x \in H$, and $H$ is a subgroup (and thus closed under multiplication), it is clear that $hxh^{-1} \in H$.
On the other hand, we have $x \in K$ as well, and since $K$ is normal, and $h \in G$ (since $H$ is a subgroup of $G$), we have $hxh^{-1} \in K$.
It follows then, that for any $x \in H \cap K$, that $hxh^{-1} \in H \cap K$, that is:
$h(H\cap K)h^{-1} \subseteq H\cap K$.
The reverse inclusion can also be shown (although some texts do not require this, because their definition of normality is $gNg^{-1} \subseteq N$ for all $g \in G$):
Suppose we have some $h \in H$. If $x \in H \cap K$, then as we saw above, $h^{-1}xh \in H \cap K$ (since $h^{-1}$ is an element of $H$ if $h$ is). Thus $h(h^{-1}xh)h^{-1} = x \in h(H\cap K)h^{-1}$ for our $h \in H$. Doing this for each $h \in H$ shows that for all such $h$, we have $H \cap K \subseteq h(H\cap K)h^{-1}$, so in all cases:
$H\cap K = h^{-1}(H \cap K)h^{-1}$ for every $h \in H$.

A small aside: once you have shown $HK$ is a bona-fide subgroup of $G$, it is trivial that $K \lhd HK$ if $K \lhd G$, since conjugation by an element of $HK$ is still conjugation by an element of $G$.

Aside #2: Note that for $HK$ to be a subgroup, we only need one of $H,K$ to be normal. This high-lights a certain asymmetry in the roles of $H$ and $K$, a wrinkle that happens because groups need not be abelian.
