Implication in Zassenhaus Lemma I need to proof Zassenhaus Lemma using the First Isomorphism Theorem and I have a problem with the following implication:

$H' \vartriangleleft H < G, K' \vartriangleleft K < G \Longrightarrow H'(H \cap K') \vartriangleleft H'(H \cap K) < H.$

I have shown that $H \cap K' \vartriangleleft H \cap K, $ but I don't see why $H'(H \cap K') \vartriangleleft H'(H \cap K)$. It should follow from the First Isomorphism Theorem and from $H' \vartriangleleft H,$ but I don't see how.
 A: Actually, the whole Zassenhaus Lemma is a straithforward application of the Second Isomorphism Theorem which states:

If $H\subseteq N_G(K),$ then $HK = KH$ and $H/(H\cap K)\cong HK/K$

(where $N_G(K)$ denotes the normalizer of $K$ in $G$, and all the appropriate subgroups can be verified to be normal)
Let me change your notation a bit (so I can relate it to 2nd isomorphism theorem easier). Let $U$ and $V$ be subgroups of $G$, $u\trianglelefteq U$, $v\trianglelefteq V$ be normal subgroups. Then Zassenhaus Lemma claims that $$u(U\cap V)/u(U\cap v)\cong (U\cap V)v/(u\cap V)v.$$
The proof uses symmetry; if we could prove that $$u(U\cap V)/ u(U\cap v)\cong (U\cap V)/(u\cap V)(U\cap v)$$
 then by symmetry we would have $$(U\cap V)v/(u\cap V)v\cong (U\cap V)/(u\cap V)(U\cap v)$$ as well, and this would be the end of the proof.
So, let us denote $H = U\cap V$ and $K = u (U\cap v)$. I leave it as an exercise (and will write it down if requested) to prove that $$H\subseteq N_G(K),$$  $$u(U\cap V) = u(U\cap v)(U\cap V),$$ and $$(U\cap V)\cap u(U\cap v) = (u\cap V)(U\cap v).$$
The rest is the 2nd Isomorphism Theorem.
Edit:
You want to prove that $K$ is normal in $HK$ (in my notation). As I noted, this is a part of the proof of the Second Isomorphism Theorem.
Let us assume that $H\subseteq N_G(K)$. Then obviously $HK = KH$ and it is  a one liner to prove that $HK$ is a subgroup of $G$: $$(hk)(h'k') = hh'\underbrace{(h')^{-1}kh'}_{\in K}k'\in HK$$ (almost, since $(hk)^{-1}=k^{-1}h^{-1} \in KH = HK$)
$K$ is normal in $HK$: 
Let $h\in H$, $k\in K$, $x\in K$,
$$(hk)x(hk)^{-1} = h\underbrace{kxk^{-1}}_{\in K}h^{-1} \in K$$ because $h\in N_G(K)$.
This answers your inquiry. 
But it still remains unanswered why $u(U\cap V)(U\cap v) = u (U\cap V)$ (i.e. why $HK =  u (U\cap V)$ in my notation). Well, both the inclusions are obvious if you observe that $U\cap v \subseteq U\cap V$, so $(U\cap V)(U\cap v) = U\cap V$.
Also, why is $U\cap V\in N_G(u(U\cap v))$? Let $x\in U\cap V$ and $u_0v_0\in u(U\cap v)$. Then $$ x(u_0v_0) = u_0'xv_0 = u_0'v_0'x$$ where $u_0'\in u$ since $u_0\in u$ and $u$ is normal in $U$ (and $x\in U\cap V$); and $v_0'\in v$, since $v_0\in v$ and $v$ is normal in $V$ (and $x\in U\cap V$). The only question is whether $v_0'\in U$, but that follows immediately from $xv_0 = v_0'x\implies v_0' = xv_0x^{-1}\in U$. This proves $x(U\cap v) = (U\cap v)x$, for any $x\in U\cap V$, and thus $U\cap V\in N_G(u(U\cap v))$.
A: Perhaps it is more clear if you prove the following:

Claim: If $C \trianglelefteq H$ and $A \trianglelefteq B \leq H$, then $CA \trianglelefteq CB \leq H$.

Then your claim follows with $C = H'$ and $A = H \cap K'$, $B = H \cap K$. Note that in this situation $A \trianglelefteq B$  since $K' \trianglelefteq K$.
Maybe you want to try to prove the claim yourself. If you get stuck, here is a proof:

 $CA$ and $CB$ are subgroups since $C$ is normal, and $CA \leq CB$ is obvious. Now $B$ normalizes $CA$ since $C$ is normal, and since $A \trianglelefteq B$. Furthermore, $C$ normalizes $CA$ since $C$ is a subgroup of $CA$. Thus $CB$ normalizes $CA$, that is, $CA \trianglelefteq CB$.

