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My question is kind of selly, but I really need to understand the consept very well, so my problem is:

le G be a torsion-free group (i.e $\quad\forall x \in G$ $x$ has an infinite order).

I know that if $H \subset G$ then it is also torsion-free, and if $N \trianglelefteq G$ then $G/N$ is not torsion-free.

I would know if $N$ and $G/N$ are both torsion- free groups, then is $G$ also a torsion-free group or not?

And if $G$ is a torsion-free group, is it of infinite order?

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    $\begingroup$ The correct definition of a torsion-free group is that every non-identity element $x \in G$ has infinite order. So the group of order $1$ is torsion-free. But all other torsion-free groups are infinite. $\endgroup$ – Derek Holt Dec 28 '15 at 18:05
  • $\begingroup$ and what about the stability by extension? $\endgroup$ – Lauren Dec 28 '15 at 18:08
  • $\begingroup$ Ii think you mean N and G/N in title $\endgroup$ – Bhaskar Vashishth Dec 28 '15 at 18:19
  • $\begingroup$ yes, ill edit it $\endgroup$ – Lauren Dec 28 '15 at 18:19
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Let us clear up a few things.

A group $G$ is torsion-free if and only if, for all $x \in G$, either $x$ is the identity or $x$ has an infinite order (beware the identity!). In particular, the trivial group is torsion-free.

Let $G$ be a torsion-free group. Either $G$ is trivial, or $G$ has an element $x$ which is not the identity. In the later case, the sequence $(x^n)_{n \in \mathbb{Z}}$ is a sequence of pairwise distinct elements of $G$, hence $G$ is of infinite order. So you have the dichotomy: either $G$ is trivial, or it has infinite order.

Let $G$ be torsion free, and $N < G$. Then $G/N$ may or may not be torsion free. For instance,

  • take $G := \mathbb{Z}$ and $N := \{0\}$. Then $G/N \simeq \mathbb{Z}$ is torsion-free.

  • take $G := \mathbb{Z}$ and $N := 2 \mathbb{Z}$. Then $G/N \simeq \mathbb{Z}/2\mathbb{Z}$ is not torsion-free.

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Finally, let me answer your question. Le $G$ be a group, $N < G$ a normal subgroup of $G$. Assume that $N$ and $G/N$ are torsion free. We want to prove that $G$ is also torsion-free.

Let $\pi : G \to G/N$, $\pi (x) = xN$ be the canonical morphism.

Let $x \in G$. If $x$ has finite order, then there exists $n > 0$ such that $x^n = e$, so $N = \pi(x^n) = \pi(x)^n$. Hence, $\pi(x)$ also has finite order in $G/N$. Since $G/N$ is assumed to be torsion-free, its only element with finite order is the identity, so $\pi(x) = N$, that is, $x \in N$. Since $N$ is also assumed to be torsion-free, its only element with finite order is $e$, so $x = e$.

Hence, the only element of $G$ with finite order is its identity, so $G$ is also torsion-free.

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  • $\begingroup$ thank you very much, it seems clear now. $\endgroup$ – Lauren Dec 28 '15 at 18:15
  • $\begingroup$ and if $G$ is torsion-free, is it of infinite order? $\endgroup$ – Lauren Dec 28 '15 at 18:16
  • $\begingroup$ @MerieMarissa No since the trivial group is torsion-free. $\endgroup$ – MCT Dec 28 '15 at 18:24
  • $\begingroup$ ok, but can we say that the trivial group is periodic also, since $1^{n}=1$ $\endgroup$ – Lauren Dec 28 '15 at 18:27
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If $x \in G$ is a torsion element then either $x \notin N$ in which case $\bar{x}$ is a torsion element in $G/N$, or $x \in N$ and $N$ is not torsion free. Additionally, if $G$ is finite, say $|G| = n$, then every $x \in G$ is a torsion element since $x^n = e$.

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  • $\begingroup$ but in the hypothese both $N$ and $G/N$ are torsion-free $\endgroup$ – Lauren Dec 28 '15 at 18:05
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    $\begingroup$ exactly. this a proof by contradiction. $\endgroup$ – basket Dec 28 '15 at 18:07

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