Torsion Subgroup

**Bernhard** · Oct 21st 2011, 03:20 AM

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 $\in$ 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 $\in$ G| |g| is less than infinity}

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

Need to show x $y^{-1}$ $\in$ H

ie need to show |x $y^{-1}$ | is less than infinity

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

Can anyone help?

Peter

**Deveno** · Oct 21st 2011, 04:31 AM

suppose $|y| = n$ so that $y^n = e$ . then $(y^{-1})^n = (y^n)^{-1}$ as you can verify by multiplication.

therefore, $(y^{-1})^n = (y^n)^{-1} = e^{-1} = e$ , so $|y^{-1}|$ is finite.

so $y^{-1} \in H$ if $y \in H$ .

now suppose that $x,y \in H$ , so that $|x| = m, |y| = n$ .

$(xy^{-1})^{mn} = x^{mn}(y^{-1})^{mn}$ (since G is abelian)

$= (x^m)^n((y^{-1})^n)^m = e^ne^m = e$ , so $|xy^{-1}| < \infty$ , that is $xy^{-1} \in H$ .

note we didn't show what the order of $xy^{-1}$ 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:

$(fg)^k(n) = n - k \neq n$ so fg is of infinite order.

**Bernhard** · Oct 21st 2011, 03:37 PM

Thanks Deveno

Really helpful - and really neat too!

Peter

**Swlabr** · Oct 22nd 2011, 11:49 AM

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

$\langle a, b; a^2=1, aba=b^{-1}\rangle$

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

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

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