Let $H$ be normal subgroup of $G$. If $G/H$ is cyclic group generated by $aH$, prove that $G=KH$ where $K=\langle a\rangle$. Let $H$ be normal subgroup of $G$. If $G/H$ is cyclic group generated by $aH$, prove that $G=KH$ where $K=\langle a\rangle $. I would like someone to check my solution.
First of i will prove that $G$ is subset of $KH$ and then that $KH$ is subset of $G$. 
Since $G/H=\{ gH | g \in G \}=\langle aH\rangle$ we have that $(\forall g \in G)$$(\forall h \in H)$$(\exists l \in \mathbb Z)$$(\exists h' \in H)$ such that $gh=ah'ah'...ah'$ ($ah'$ multiplied $l$ times by itself).
Since $H$ is normal subgroup it follows that $Ha=aH$ i.e. $h'a=ah''$ (for some $h''$ from $H$). So now we have   $gh=aah''h'ah'...ah'$ Since $H$ is normal subgroup it follows that $HH=H$ i.e. $h''h'=h'''$ (for some $h'''$ from $H$). So $gh=aah'''ah'...ah'$. And so on until i get $gh=aa...h^*=a^lh^*$ where $h^* \in H$.
Now we multiply left and right sides by $h^{-1}$ from right side. Now we get $ghh^{-1}=g=a^lh^{*}h^{-1}=a^lh^{**}$ where  $a^l \in K$ and $h^{**}$ is in $H$. 
In other words i can represent $\forall g \in G$ as product of some element in $K$ and some element in $H$ so it follows that $G$ is subset of $KH$.
Now since $H$ is subgroup and $\langle a\rangle$ is cyclic group generated by $a \in G$ (i'm assuming this, nowhere in formulation of problem is this said, but it wouldn't make any sense if $a$ is not in $G$), and $G$ is group i.e. it is closed for binary operation, it follows that every element in $KH$ is some $g$ in $G$ i.e. $KH$ is subset of $G$.
Therefore we have that $G=KH$
 A: Your proof works, but the proof of the first inclusion can be made much shorter if you use the hypothesis that $G/h=\langle a \rangle$ a little differently.

Since $G/H=\{ gH | g \in G \}=\langle aH\rangle$ we have that $(\forall g \in G)$$(\forall h \in H)$$(\exists l \in \mathbb Z)$$(\exists h' \in H)$ such that $gh=ah'ah'...ah'$ ($ah'$ multiplied $l$ times by itself).

Here it's easier to notice that $G/H=\langle aH \rangle$ implies  that $gH = a^kH$ for some $k \in \Bbb Z$, so $g\in a^kH$, meaning $g=a^kh$ for some $h \in H$. 
Then you don't have to move the $a$ and the elements of $H$ around like you did in the next paragraph.
A minor point:  

$HH=H$ i.e. $h''h'=h'''$ (for some $h'''$ from $H$). 

This is true for any subgroup, not just for normal ones (as you noted, it follows from closure, and all subgroups are closed)
A: Here is how I see it: the cosets $a^kH$ partition $G$, so we immediately have that any $g \in G$ is in some coset $a^kH$, so $g = a^kh$ for some $k \in \Bbb Z^+$ and some $h \in H$, that is: $g \in KH$ (since $a^k \in K$).
Thus $G \subseteq KH$.
But, obviously, $KH \subseteq G$, so the two are equal.
(it might seem mystifying that I did not use normality of $H$: well, I did, but it's "buried" in the fact that any coset is of the form $a^kH$, which we have from the fact that $G/H = \langle aH\rangle$ which only makes sense if $G/H$ is, in fact, a group, which means $H$ must be normal).
