Let H be a normal subgroup of G.
$1$. Assume that H is a normal subgroup of G. Show the product $aHbH$ is independent of the representative chosen from $aH$ and $bH$, so show that $(ah)(bh') \in abH$.
$2$. Show that the order of the coset $aH$ must divide the order of $|a|$ in G. So show that $\frac{|a|}{|aH|}$.
$1$. Let $ah_1 \in aH$ and $bh_2 \in bH$. Since H is normal subgroup of G, $ah_1bh_2=abh_3h_2$. But this gets me no where since I have to show that the representatives are independent. Similarly, $abh \in abH$, but its tough to decompose.
$2$. Where would I even begin to show this. Other then Since the order of the coset must divide the order of group and the order of the element must also divide the order of the group. How do these two relate? If the two numbers were relatively prime then things would be different, but we dont know that.
Best Answer
For #1:
Let $ah \in aH, bh' \in bH$. Then since $H$ is normal, $bh'b^{-1} \in H$, i.e. $h' = b^{-1}h''b$ for some $h''\in H$. Hence, $bh' = h''b$. Then $ahbh' = ahh''b$. Since $H$ is a subgroup $hh''= k$ for some $k \in H$. Finally, we use normality once again to note that $b^{-1}kb = k'$ for some $k' \in H$, i.e. $kb = bk'$. But then $ahbh' = akb = abk'$. So $ahbh' \in abH$.
For #$2$:
The order of the coset $aH$ is the least integer $n$ so that $(aH)^{n} = H$, the identity coset. Using part #1, show that $(aH)^{n} = a^{n}H$ via induction. Then, note that $(aH)^{n} = H$ if and only if $a^{n} \in H$, so $n$ is the least integer such that $a^{n}$ is in $H$. Since $H$ is a subgroup, it contains the identity element $e$. Thus, if $|a|=k$, then $a^{k} = e \in H$. Thus, $(aH)^{k} = eH = H$, so $|aH|$ divides $|a|$.