0

We know an abelian group has a subgroup which all elements in it has finite order . Now we remove condition abelian group , the clause is that true? If it's false give an example.

Metin Y.
  • 1,464
Hung nguyen
  • 699
  • Hint: The subgroup that works in the abelian case still works. – Tobias Kildetoft Apr 03 '13 at 13:48
  • @Tobias: what do you mean? The product of two finite-order elements is not necessarily of finite order. – Martin Apr 03 '13 at 13:53
  • 1
    @Martin I may have misunderstood the OP. But as it is currently stated, he is not asking whether the set of elements of finite orders form a subgroup, but whether there is some subgroup where all elements have finite order. – Tobias Kildetoft Apr 03 '13 at 13:55
  • @Tobias: Ah. I read the question to mean whether the torsion elements form a subgroup. Of course, the cyclic subgroup generated by a torsion element is a torsion subgroup. Thanks. – Martin Apr 03 '13 at 13:56
  • 2
    The question seems to be seriously ill-posed: what clause is the OP asking about? Apparently Martin's right and the OP actually meant to ask whether the subset of all elements with finite order is a subgroup...Hard to tell what the question is, so I expect some editing and explaining or else I'll vote to close it. – DonAntonio Apr 03 '13 at 14:12
  • The integers under addition form an abelian group, but the only element with finite order is the identity, so the only subgroup with elements of finite order is the trivial subgroup. I suspect the OP is really asking something else; clarification is needed. – Josh B. Apr 03 '13 at 14:19
  • Seethis link http://math.stackexchange.com/q/279199/8581 – Mikasa Apr 03 '13 at 14:49

3 Answers3

1

A counter-example: the free product of two groups of the order 2 (more generally: of groups of finite orders).

Boris Novikov
  • 17,470
1

Consider the trivial subgroup.

Cameron Buie
  • 102,994
0

I think you just want torsion-free group: http://en.wikipedia.org/wiki/Torsion_%28algebra%29

Easy
  • 4,485