If $H \trianglelefteq G$ and $H \subseteq G' \leq G$ then $(G' \trianglelefteq G) \Leftrightarrow (G'/H \trianglelefteq G/H)$ I tried to prove the statement in both directions at once, as a chain of equivalent statements:

  
*
  
*$G' \trianglelefteq G$. 
  
*$xG'x^{-1} = G'$ for all $x \in G$.
  
*$\{xyx^{-1} \mid y \in G'\} = \{y \mid y \in G'\}$ for all $x \in G$.
  
*$\{xyx^{-1}H \mid y \in G'\} = \{yH \mid y \in G'\}$ for all $x \in G$.
  
*$\{xHyHx^{-1}H \mid y \in G'\} = \{yH \mid y \in G'\}$ for all $x \in G$.
  
*$(xH)(G'/H)(x^{-1}H) = G'/H$ for all $xH \in G/H$.
  
*$G'/H \trianglelefteq G/H$.

Here, 1 ⇔ 2 ⇔ 3 by definition, 3 ⇒ 4 by applying $\alpha \mapsto \alpha H$ to everything in the set, 4 ⇔ 5 by normality of $H$ in $G$ (we can twiddle the cosets around: $xHyHx^{-1}H=xyx^{-1}HHH=xyx^{-1}H$), and 5 ⇔ 6 ⇔ 7 by definition.
However, I’m not sure how to motivate that 4 ⇒ 3, to finish the proof. I haven't yet used the condition that $H \subseteq G'$, so it’s probably relevant. But I see no way to use it. Can I get a hint?
 A: Hint: 4 means that every $xyx^{-1}H$, where $x\in G$ and $y\in G'$, there exists $y'\in G'$ such that $xyx^{-1}H=y'H$. But what you need is $xyx^{-1}$, not the hole coset of $H$. Let $e$ denote the identity element of $G$. Of course $e\in H$. So $xyx^{-1}=xyx^{-1}e\in xyx^{-1}H=y'H$. Indeed you'll need the assumption $H\subseteq G'$ here.
P.S: I find proofs of equivalences by chain of equivalences splendid!
A: I will write $L$ instead of $G'$.
We know that 
$$H \trianglelefteq G$$
$$H \le L \le G$$
$$xLx^{-1}H = LH$$
and we need to prove
$$xLx^{-1} = L$$
Notice that $LH = L$ and that $$xLx^{-1}H = xL(x^{-1}Hx)x^{-1} = xLHx^{-1} = xLx^{-1}$$
Thus $xLx^{-1} = xLx^{-1}H = LH = L$.
A: Assert indirectly that statement 3 is not true. Then $∃y ∈ G', ∃x ∈ G$ such that $xyx^{-1} \not\in G'$. (The set on the right hand side will always be contained in the one on the left. Just take $x=e$)
Now consider $G'/H$ as a subgroup of $G/H$. Assert there exists a $z ∈ G'$ such that the left cosets $xyx^{-1}H = zH$ coincide. This is equivalent to $xyx^{-1}z^{-1} ∈ H \subseteq G'$. Now since $z ∈ G'$, $xyx^{-1} ∈ G'$, as well. A contradiction.
