Thread: Torsion Subgroup

Dummit and Foote Section 2.1 (Definitions and Examples of Subgroups) Exercise 6 reads as follows:

Let G be an abelian group. Prove that H = {g G| |g| is less than infinity} is a subgroup of G (called the torsion subgroup of G). Give an explicit example where this set is not a subgroup when G is not abelian.

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H = {g G| |g| is less than infinity}

Let x,y H ie |x| is less than infinity and |y| is less than infinity

Need to show x H

ie need to show |x | is less than infinity

? but how does one do this? Also ? explicit example ?

Can anyone help?

Peter

suppose so that . then as you can verify by multiplication.

therefore, , so is finite.

so if .

now suppose that , so that .

(since G is abelian)

, so , that is .

note we didn't show what the order of actually was, just that it was finite (and less than or equal to mn).

finding a counter-example is a bit tricky. let's use G = Sym(Z), the group of all bijections of Z.

define f(n) = -n. it's clear f is a bijection, and f has order 2.
define g(n) = 1-n. it's also clear g is a bijection, and:

g(g(n)) = g(1-n) = 1 - (1-n) = 1 - 1 + n = n, so g also has order 2.

but fg(n) = f(g(n)) = f(1-n) = n-1, and for any positive integer k:

so fg is of infinite order.

Thanks Deveno

Really helpful - and really neat too!

Peter

Another counter-example is the infinite dihedral group. This is the group with presentation,



so it is basically the regular dihedral group but the rotation element has infinite order.

So, notice that this group is generated by and , both of which are of order 2. Thus, the torsion elements generated the entire group, and so do not form a subgroup!