Let $G$ be a finite group and let $H$ be a normal subgroup of $G$. Prove $|gH| \in G/H$ divides $|g|\in G$. Let $G$ be a finite group and let $H$ be a normal subgroup of $G$.  Prove that the order of the element $gH$ in $G/H$ must divide the order of $g$ in $G$.
I see that if we have: $H = \{e, h_1, ..., h_2\}, gH = \{g, gh_1, ..., gh_2\}, ..., g^{k-1}H = \{g^{k-1}, g^{k-1}h_1, ..., g^{k-1}h_2\}$, then the only true deciding element in $H$ in $e$, because no matter what $g$ you use you will produce the original coset once $g^{|g|} = e$.
But this seems to show that $|g| = |gH|$.
Am I missing something here?  Because this does divide it, but only trivially.
 A: By the division algorithm, we can write, $|g| = q\cdot |gH|+r$ for integers $q,r$ where $0 \leq r < |gH|$. Notice that $g^{|g|} = e$, so
$$(gH)^{|g|} = (g^{|g|})H = eH = H$$
and
$$(gH)^{|g|} = (gH)^{q\cdot|gH|+r} = \left((gH)^{|gH|}\right)^q \cdot (gH)^r = H^q \cdot (gH)^r = (gH)^r,$$
thus,
$$(gH)^r = H.$$
Since $H$ is the identity element of $G/H$, then the above implies that if $r$ is nonzero, then $r$ is at least the order of $gH$; however, from before we have that $r < |gH|$, thus the only possible value of $r$ is $0$. Hence, $|g| = q \cdot |gH|$, thus $|gH|$ divides $|g|$.
A: Writing $g^{\lvert g\rvert}H=H$ only means $\lvert g\rvert$ is a multiple of the order of $gH$.
Indeed we very well may have $g^rH=H$ for smaller values of $r$, since $g^rH=H$ means $g^r\in H$, not $g^r=e$.
As a counterexample, consider the cyclic group $C_4$, $g $ a generator, $H$ the subgroup $\langle\mkern 1mu g^2\rangle$. $G/H$ is the cyclic group $C_2$, generated by $gH$. So $\lvert gH\rvert=2$, while $\lvert g\rvert=4$.
A: The equality is not true in general because if $g\in H\setminus\{e\}$ then $gH=H$ but the order of $g$ is not $1$. 
To take a concrete example, let $(G,\circ)$ be $(\mathbb{Z}_2 \times \mathbb{Z_2},+)$ where the "addition" is done component wise. Let $H:=\{(0,0),(1,1)\}$. Now observe that, $$(1,1)+H=H$$ and hence $\lvert (1,1)+ H\rvert=1$ but $|(1,1)|=2$.  
