What about my proof is "nonsense"? I am working on a question from Fraleigh's "A First Course In Abstract Algebra":

A torsion group is a group all of whose elements have finite order. A
  group is torsion free if the identity element is the only element of
  finite order. A student is asked to prove that if $G$ is a torision
  group, then so is $G/H$ for every normal subgroup $H$ of $G$. The
  student writes:

We must show that each element of $G/H$ is of finite order. Let $x \in G/H$

  
  
*
  
*Why does the instructor reading this proof expect to find nonsense
  from here on in the students proof? 
  
*What should the student have
  written? 
  
*Complete the proof.
  

So I started thinking and just thought of point 3:
We must show that each element of $G/H$ is of finite order. 
Let $x \in G/H$.
Observe that $x \in G$ as $G/H \leq G$, but since $G$ is a torsion group, and $x$ is in $G$, $x$ must have have finite order. Q.E.D.
This seems fine to me, but I think I am doing something silly since, the question leads me to believe so.
 A: As stated in other answers, $G/H$ is not a subgroup (or even a subset) of $G$. That being said, I object to the "nonsense" assumption. If I were writing this proof, it would look like this:
Let $x\in G/H$. Then $x=gH$ for some $g\in G$. Since $G$ is a torsion group, $\lvert g\rvert=n$ for some $n<\infty$; in particular, $g^n=e$.
Then \begin{align*}x^n&=(gH)^n\\ &=g^nH\\&=eH\\&=H\end{align*}
So $\lvert x\rvert \leq n<\infty$. Thus every element of $G/H$ has finite order, i.e. $G/H$ is a torsion group.
A: The proof should start: "Let $x \in G$..."
Why? Because $G$ is the group that has KNOWN properties, and we (as yet) no NOTHING about $G/H$ (although we ought to be proving something about it, shortly).

EDIT: There is no reason to suppose that $G/H$ is even isomorphic to a subgroup of $G$. For example, let:
$G = \{1,-1,i,-i,j,-j,k,-k\}$ where $ij = k, jk = i, ki = j$. Let $H = \{1,-1\}$.
Then $G/H = \{H,iH,jH,kH\} \cong V$, but $G$ has only $1$ element of order $2$, so cannot be isomorphic to $V$, which has $3$ elements of order $2$.
A: The issue I had was  thinking about the set $G/H$. As stated by David Wheeler, the $G/H$ is not necessarily isomorphic to $G$. 
Although I didn't assume it was isomoprhic while reasoning about it, I did not think of $G/H$ as $\{H, iH, jH, kH\}$, I instead thought of it as a particular instance of this. So $G/H = gH$ for some $g \in G$ was the way I "pictured it".
Instead I should of though of $G/H$ as a set of sets. And made the proof like so:
Let $x \in G$ with order $n$, Then we wish to show that $xH$ has finite order. Notice the identity element is $H$. We have: $(xH)^2 = (xH)(xH) = x^2H$. It is easy to see that:
$$(xH)^n = x^nH = eH = H$$
As desired.
