The only subgroup of finite order of a an infinite group is the identity
My proof
Let G be a group with infinite order
let H be a subgroup and letand
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which make H with an infinite order
am I rite ?
is there any counter example
The only subgroup of finite order of a an infinite group is the identity
My proof
Let G be a group with infinite order
let H be a subgroup and letand
![]()
![]()
which make H with an infinite order
am I rite ?
is there any counter example
More generally, take any fieldthen if there exists non-zero
with
(i.e. a non-trivial
root of unity) then the group
of
-roots of unity form a finite subgroup of
. This clearly generalizes Dr. Revilla's example (so that, for example, the
-roots of unity in
).
But, perhaps even more of a trivial counterexample. Take any infinite groupand consider
....
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