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It is known that the torsion subset of a group is not, in general, a subgroup. For example, consider the infinite dihedral group $\mathcal{D}_\infty = \left < x, y \mid x² = y² = 1 \right >$.

Is the subset of non-torsion elements "always" a subgroup, or can we have scenarios where two elements $x,y$ have infinite order whereas their product $xy$ has only finite order?

Intuitively, I would say no, since I cannot think of any counterexamples.

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A trivial example: $x$ can have infinite order but $xx^{-1}$ has finite order. Slightly less trivially, one can look at something like $\mathbb{Z}\oplus\mathbb{Z}_2$ where (1,1) and (-1,0) have infinite order but their sum has finite order. –  Miha Habič Aug 18 '13 at 16:11
    
Thanks for the other example. I was already trying to construct some elements in matrix groups. –  yannickvda Aug 18 '13 at 16:19
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up vote 1 down vote accepted

In general, the sets of finite and infinite elements are each closed under conjugacy but will not be subgroups. So take these sets for $H$.

The set of infinite elements will never form a subgroup. However, if you attach the identity it sometimes will and sometimes will not. For example,

  • Take $C_2\ast C_3\cong \langle a, b; a^2, b^3\rangle$. Then the set of elements of infinite order is closed under conjugacy, but it is not a subgroup as $ab$ and $ba$ both have infinite order (erm...I cannot think of a simple reason why - it follows from the theory of free products) while $ba\cdot ab=ba^2b=b^2$ has finite order.

  • Take $D_{\infty}\cong \langle a, b; a^2, b^2\rangle$. Then every element of infinte order is in the infinite cyclic subgroup $\langle ab\rangle$, and so the set of elements of infinite order along with the identity forms a normal subgroup of $G$.

For the case of torsion elements, again both cases are possible:

  • Take $D_{\infty}\cong \langle a, b; a^2, b^2\rangle$. Then the set of elements of finite order is closed under conjugacy, but it is not a subgroup as $ab$ has infinite order.

  • Take a group $G$ which is the cross product of a finite group with a torsion-free group (every element has infinite order), for example $\mathbb{Z}\times A_5$. Then the set of elements of finite order forms a normal subgroup of $G$.

As a nod to your comment on the main question, note that all the examples I have given are "matrix groups", that is, you can view then as matrices (over $\mathbb{Z}$).

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Take a group G which is the cross product of a finite group with a torsion-free group. [...] Then the set of elements of infinite order along with the identity forms a normal subgroup of G. What if you take $\Bbb Z\times C_2$. Then the infinite order elements along with the identity are all except $(0,1)$ –  Stefan Hamcke Aug 18 '13 at 17:05
    
@StefanH. Yes, sorry, I clearly didn't think that through properly! However, $D_{\infty}$ works. I'll correct it. –  user1729 Aug 18 '13 at 17:52
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