TITLE: What about my proof is "nonsense"?
QUESTION [6 upvotes]: 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.

REPLY [5 votes]: 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.