A rich question regarding subgroups, normal subgroup, and subgroup generated by a set. Let $G$ be a group of identity element $e$.
Let $D(G)$ be the subgroup generated by $P$ where 
$P$ = {$xyx^{-1}y^{-1};$ ${x,y}\in G$ }.
Show that:
a) $D(G)$= {$e$} if $G$ is abelian.
b) $D(G)$ is a normal subgroup of G.
c) $G/D(G)$ is abelian.
d) If $H$ is a normal subgroup of $G$ and $G/H$ is abelian, then $D(G)\subseteq H$.
e) If H is a proper subgroup of G and $D(G)\subseteq H$, then H is normal and $G/H$ is abelian.
Regarding (a):
If $D(G)$= {$e$}, then $ \forall {x,y}\in G$, $xyx^{-1}y^{-1}$ = $e$, since any finite product of elements of $P\cup P^{-1}$ will be equal the identity that means that all the elements are the identity. So $xy$ = $yx$.
Conversely, If $xy$ =$yx$ $ \forall {x,y}\in G$, then $xyx^{-1}y^{-1}$ = $e$ and then  $D(G)$= {$e$}.
Regarding (b):
I know that I should prove that $aza^{-1}\in D(G)$ for some $z\in D(G)$ and that $z= t_1t_2t_3....t_s$ for $s\geq 0$. I tried alot but didn't reach.
Regarding (c):
I thought that $G/H$ is abelian if $G$ is, for $H$ a normal subgroup in $G$. But here this is not the case. My problem is that I don't have the ability even to start! I'm really afraid of my exam!.
No ideas for (d) and (e). Please help me with hints, examples, or even theorems I may not know. Thanks in advance and sorry for this long question! 
 A: You may (or may not) find the following notation for commutators handy:
Write $[x,y] = xyx^{-1}y^{-1}$ (some people write the inverses first, it's not that important, here).
Write $a^g = g^{-1}ag$ (here, it is important to put the $g^{-1}$ first, because we want the following to be true:
$a^{gh} = (a^g)^h$).
Then (b) is a consequence of the fact that: $[x,y]^g = [x^g,y^g]$ (this basically says that the map $z \mapsto z^g$ is an automorphism of $D(G)$, for every $g \in G$, which is another way of saying $D(G) \lhd G$).
(c) is relatively straightforward: we need to show that:
$(D(G)x)(D(G)y) = (D(G)y)(D(G)x)$ for every pair of cosets $D(G)x,D(G)y \in G/D(G)$.
By the very definition of coset multiplication, we have:
$(D(G)x)(D(G)y) = D(G)(xy)$, and $(D(G)y)(D(G)x) = D(G)(yx)$, so it will suffice to show:
$D(G)(xy) = D(G)(yx)$. That is true, if and only if:
$xy(yx)^{-1} \in D(G)$. But $xy(yx)^{-1} = xyx^{-1}y^{-1} = [x,y] \in D(G)$, by the definition of $D(G)$.
(d) is "meatier": it does not seem that obvious from the outset that if $H \lhd G$ and $G/H$ is abelian, that $D(G) \leq H$ (here, and below, I use "$\leq$" to mean "is a subgroup of").
But the proof for (c) gives us a way to proceed: for any $x,y \in G$, we have (since $G/H$ is abelian) that:
$Hxy = (Hx)(Hy) = (Hy)(Hx) = Hyx$. Can you continue?
(e) is like the "flip side" of the coin: suppose $D(G) \leq H$. The hard part is showing $H$ is normal.
But look: we know that for any $g \in G$, that $g^{-1}hgh^{-1} = [g^{-1},h] \in D(G)$, and thus in $H$.
It then follows that $(g^{-1}hgh^{-1})h = g^{-1}hg \in Hh = H$, so $H \lhd G$. I leave the rest of (d) in your capable hands.
A: Regarding (b):
It's enough to prove that $$axyx^{-1}y^{-1}a^{-1} \in D(G)$$
$axyx^{-1}y^{-1}a^{-1}=axa^{-1}aya^{-1}ax^{-1}a^{-1}ay^{-1}a^{-1}=axa^{-1}aya^{-1}(axa^{-1})^{-1}(aya^{-1})^{-1} \in D(G)$
Regarding (c) and (d):
$G/H$ being abelian is not the same as $G$ being abelian. Try to prove: $G/H$ is abelian iff $D(G) \subseteq H$. It's not hard, it follows from the definition.
