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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 $x \in G/H$

  1. Why does the instructor reading this proof expect to find nonsense from here on in the students proof?
  2. What should the student have written?
  3. 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.

Dair
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    What does it mean to say that $x \in G$? $G/H$ is not a subgroup of $G$, it's a quotient. I guess Fraleigh is objecting because the proof is really going to take place in $G$, not $G/H$; let $\varphi: G \to G/H$ be the quotient map. $\varphi$ is surjective. For every $x \in G$, $x^n = 1_G$ for some $n$; so $\varphi(x)^n = 1_{G/H}$, and so every element of $G/H$ is torsion. But one could easily say "let $x \in G/H$, then suppose $y \in G$ has $\varphi(y) = x$", so I don't really see the point. –  Mar 18 '15 at 02:55
  • @MikeMiller: I was thinking, that $x = ah$ for some $a \in G$ and $h \in H (\leq G)$, by closure of multiplication it has to be a member of $G$? Does this not make that much sense? – Dair Mar 18 '15 at 03:01
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    Yes but when you take $x$ to be an element of the quotient group, $x$ becomes a subset of the group. I remember Fraleigh using this scheme a few times throughout the book in questions like this. The "nonsense" usually lies in that an element of $G/H$ is of the form "$xH$", not "$x$" (which, one might argue, does not necessitate nonsense, as the next statement could be: "Then $\exists g\in G: x=gH$"). – Alp Uzman Mar 18 '15 at 06:31
  • @Uzman I agree with your breakdown. Since $G/H$ is a group, it has well-defined elements. As such, $x\in G/H$ is a perfectly sensible thing to say and makes more sense than letting $x\in G$ (because we aren't interested in showing anything about elements of $G$). It will, of course, not be helpful until $x$ is defined in terms of a coset, but I think it is actually a better way to write the proof. – rnrstopstraffic Mar 18 '15 at 20:07

3 Answers3

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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.

rnrstopstraffic
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    And this is fine. The question isn't posed fairly, one can "work one's way out of the trap". As pointed out in a comment above, one could also write: "Let $x \in G/H$. Then $x = \pi(g)$ for the canonical projection homomorphism $\pi: G \to H$, since this is surjective, Hence..." – David Wheeler Mar 18 '15 at 23:55
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    Sure, although I would say that it isn't a trap to begin with; it's the more appropriate way to approach the problem (starting with an element in the group about which we want to prove something). Also, while the canonical projection approach is entirely accurate, given the textbook and level of OP, it's probably not the most accessible route. – rnrstopstraffic Mar 19 '15 at 14:34
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    It's sort of a point-of-view thing. We are going to have to work back in $G$ either way, the fact that the cosets $gH$ partition $G$ is equivalent to $g\mapsto gH$ being surjective (something we NEED to find some element in $G$ we can leverage). A lot of people find cosets counter-intuitive, but the idea of images of functions don't seem so bad. Either way, we need $g^n = e \implies (gH)^n = H$ (or: $g^n = e \implies [\pi(g)]^n = \pi(e)$, either way). The important thing is that the order of a torsion element in $G$ gets "knocked down" (or stays the same) in the quotient group. – David Wheeler Mar 19 '15 at 14:47
  • Well said. Also, when I look at the section in Fraleigh which references this, (Section 14), it is immediately preceded in the section by the Fundamental Homomorphism Theorem, so the projection should be a fairly comfortable notion at that point. – rnrstopstraffic Mar 19 '15 at 16:11
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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 $2$.

David Wheeler
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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.

Dair
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