Properties of normal subgroups If $K$ is a normal subgroup of $H$ then is it true that $KN$ is a normal subgroup of $HN$ where $H$ is a subgroup of $G$ and $N$ is the normal subgroup of $G$? If not, then what conditions do we require for this?
 A: Yes, this is true. 
Hints.
Check that if you conjugate any element of $KN$ by an element of $hn$, then it remains in $KN$. Do this by checking it first for $n$, and then for $h$, and deducing it for $hn$. 
Relevant general facts are that if $G$ is any group, and $H$ and $N$ are subgroups, with $N$ normal, then:
(1) $HN$ is a subgroup of $G$. 
(2) Any element of the form $hn$, with $h \in H$, $n \in N$, can also be written as $n'h$ for some $n' \in N$. 
A: Your assumptions are $K \triangleleft H$, $H \leq G$ and $N \triangleleft G$ and you are asking whether $K\cdot N \triangleleft H\cdot N$.
The answer is that $K\cdot N$ is normal in $H \cdot N$ and one way to show this is to simply show that it is invariant under conjugation with elements of $H\cdot N$. Let $h\cdot n \in H\cdot N$ and $k \cdot n' \in K \cdot N$ (with $n, n' \in N$, $h \in H$ and $k \in K$). Then 
$$ (h \cdot n) \cdot (k \cdot n') \cdot (h \cdot n')^{-1} = h \cdot n \cdot k \cdot n' \cdot n^{-1} \cdot h^{-1}.$$
Since $N \triangleleft G$ and $k \in G$ we have $n \cdot k = k \cdot n''$ for some $n'' \in N$. Similarly, since $H \triangleleft K$, $h \cdot k = k' \cdot h$ for some $k' \in K$. We thus get
$$h \cdot n \cdot k \cdot n' \cdot n^{-1} \cdot h^{-1} = h \cdot k \cdot n'' \cdot n' \cdot n^{-1} \cdot h^{-1} = k' \cdot (h \cdot n'' \cdot n' \cdot n^{-1} \cdot h^{-1}).$$
Since $N \triangleleft G$ and $h \in H \leq G$, $h \cdot (n'' \cdot n' \cdot n^{-1}) \cdot h^{-1} \in N$. Since $k' \in K$ we get
$$k' \cdot (h \cdot n'' \cdot n' \cdot n^{-1} \cdot h^{-1}) \in K \cdot N$$
as needed.
A: A conceptual way of seeing this, which makes use of the third isomorphism theorem, is the following.
Consider the epimorphism $\pi : G \to G/N$ that maps $g \mapsto g N$. 


*

*If $\mathfrak{K} \triangleleft \mathfrak{H} \le G/N$, then $\pi^{-1} (\mathfrak{K}) \triangleleft \pi^{-1} (\mathfrak{H}) \le G$. This is because if $\pi(k) \in \mathfrak{K}$ and $\pi(h) \in \mathfrak{H}$, then $\pi(h^{-1} k h) = \pi(h)^{-1} \pi(k) \pi(h)  \in \mathfrak{K}$. (Here $\pi^{-1}(A) = \{ g \in G : \pi(g) \in A \}$ for $A \subseteq G/N$.)

*If $K \triangleleft H \le G$, then 
$$\frac{KN}{N} = \pi(K) \triangleleft \pi(H) = \frac{HN}{N} \le \frac{G}{N}.$$
This is because if $h \in H$ and $k \in K$, then $\pi(h)^{-1} \pi(k) \pi(h) = \pi(h^{-1} k h) \in \pi(H)$.


Now if $K \triangleleft H \le G$, apply the two arguments to get that
$$
KN = \pi^{-1}(\pi(K)) \triangleleft \pi^{-1}(\pi(H)) = HN \le G.
$$
