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If $G$ is abelian, $T(G)$ is the torsion subgroup, then $G/T(G)$ is torsion free.

If $T(G) = \{1\}$, then $G$ is called a torsion-free group. Below is what I did to prove this statement.

$G/T(G) = \{T(G)g\mid g\in G\}$. If $g \in G$, then $T(G)g = T(G)$, otherwise $T(G)g = G - T(G)$. $So T(G/T(G))= \{T(G)\}$. $1$ is in $T(G)$ for sure, but it's also possible that there exists other elements in $T(G)$.

Could anybody point out where I am wrong?

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"If $g \in G$" is probably not what you meant. –  Dylan Moreland Mar 15 '12 at 3:17
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2 Answers

up vote 2 down vote accepted

First: it is false that if $g\in G$ then $T(G)g = T(G)$, but I suspect you meant "$g\in T(G)$", which would make that statement true.

Second: it is false that if $g\notin T(G)$ then $T(G)g=G-T(G)$. It's true that the coset $T(G)g$ is disjoint from $T(G)$, and hence is contained in $G-T(G)$, but it need not be equal to $G-T(G)$. In general it won't be (it will only be the case if $T(G)$ is of index $2$).

Here's what you need to do: take an element of $G/T(G)$, which looks like $T(G)g$ with $g\in G$, and assume that it is a torsion element: that is, there exists $n\gt 0$ such that $(T(G)g)^n = T(G)$. This will occur if and only if $T(G)g^n = (T(G)g)^n = T(G)$, if and only if $g^n\in T(G)$.

Can you now prove that in fact $g\in T(G)$? If so, then $T(G)g = T(G)$, and you will have proven that:

If $T(G)g\in T(G/T(G))$, then $T(G)g = T(G)$.

which will show that $T(G/T(G)) = \{T(G)\}$ (which happens to be the identity of $G/T(G)$).

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Is the group of rational numbers a torsion free group? –  Shannon Mar 15 '12 at 4:55
    
@Shannon: You tell me: the rational numbers are a group under addition, so an element $a\in\mathbb{Q}$ is torsion if and only if there exists $n\gt 0$ such that $\underbrace{a+\cdots+a}_{n\text{ summands}} = 0$. Is the only torsion element $0$, or are there other torsion elements? –  Arturo Magidin Mar 15 '12 at 5:00
    
sorry, I mean the rational numbers under multiplication. –  Shannon Mar 15 '12 at 5:04
    
@Shannon: The rational numbers are not a group under multiplication (zero has no inverse). For nonzero rationals, you have two possibilities: the positive rationals under multiplication, or the nonzero rationals under multiplication. Again the question is: suppose that $q$ is a nonzero [positive] rational, and $n\gt 0$ a positive integer. If $q^n=1$, does that mean that $q=1$, or not? –  Arturo Magidin Mar 15 '12 at 15:27
    
I think the rational numbers under addition is a torsion free group. And for nonzero rational numbers under multiplication is a torsion free group too. Pls correct me if I am wrong. Thank you so much. –  Shannon Mar 15 '12 at 16:54
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It's not too hard to see that $G/T(G)$ is torsion free,

The idea is that we're sending all the torsion elements to $0$ by modding out by $T(G)$, so of course there are none left after we do that!

To put it a little more formally

Suppose $a\in G/T(G)$ is torsion. Then $a^{k} = 0$ for some $k$, i.e. $a^k \in T(G)$. But this means that $a^k$ is torsion in $G$; but of course this menas that $a$ itself was torsion, so $a\in T(G)$ i.e $a = 0$ in $G/T(G)$. Thus we see that $G/T(G)$ is torsion free.

As for your proof the statement "If $g\in G$ $T(G)g = T(G)$, otherwise $T(G)g = G - T(G)$" is certainly false.

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