finite order.

**poorna** · May 20th 2009, 09:17 AM

Can you please help me with this?
G is a group such that the intersection of all its subgroups which are different from (e) is a subgroup different from (e). Prove that every element in G has finite order.

HOw in general can we prove that a subgroup or element has finite order?

**TheAbstractionist** · May 20th 2009, 10:03 AM

Originally Posted by poorna

Can you please help me with this?
G is a group such that the intersection of all its subgroups which are different from (e) is a subgroup different from (e). Prove that every element in G has finite order.

HOw in general can we prove that a subgroup or element has finite order?

Hi pooma.

I recently proved this in one of NonCommAlg’s Fun algebra problems.

. Let $x$ be a nontrivial element in the intersection of all the nontrivial subgroups. If $x^2=e$ then $x$ has finite order. Otherwise $x \in \langle x^2 \rangle$ and so $x=x^{2n}$ for some $n\in\mathbb Z$ and so $x^{2n-1}=e$ and so again $x$ has finite order.

Suppose $x$ has order $p.$ Now let $g$ be any element in the group $G.$ If $g=e$ then $g$ has finite order. Otherwise $x$ is in the subgroup generated by $g$ and so $x=g^m$ for some integer $m.$ It follows that $g^{mp}=x^p=e$ so that $g$ has finite order.

NB: You can go further that the order of $x$ must be a prime and that every nonidentity element of $G$ has order a power $p.$

**Opalg** · May 20th 2009, 10:08 AM

Originally Posted by poorna

Can you please help me with this?
G is a group such that the intersection of all its subgroups which are different from (e) is a subgroup different from (e). Prove that every element in G has finite order.

Suppose that G has an element a of infinite order. What can you then say about the sequence of subgroups generated by $a^2,\ a^3,\ a^4,\ldots\,?$

**poorna** · May 25th 2009, 05:47 AM

ah thanks a lot!

**Swlabr** · May 25th 2009, 06:11 AM

Originally Posted by TheAbstractionist

Hi pooma.

I recently proved this in one of NonCommAlg’s Fun algebra problems.

. Let $x$ be a nontrivial element in the intersection of all the nontrivial subgroups. If $x^2=e$ then $x$ has finite order. Otherwise $x \in \langle x^2 \rangle$ and so $x=x^{2n}$ for some $n\in\mathbb Z$ and so $x^{2n-1}=e$ and so again $x$ has finite order.

Suppose $x$ has order $p.$ Now let $g$ be any element in the group $G.$ If $g=e$ then $g$ has finite order. Otherwise $x$ is in the subgroup generated by $g$ and so $x=g^m$ for some integer $m.$ It follows that $g^{mp}=x^p=e$ so that $g$ has finite order.

NB: You can go further that the order of $x$ must be a prime and that every nonidentity element of $G$ has order a power $p.$

Hi,

Just thought I'd add that both the torsion and non-torsion elements generate normal subgroups with trivial intersection, as $o(g) = o(g^h)$ for all $g,h \in G$ . Thus, the group consists of either all finite or all infinite elements, and so once you have found an element of finite order all elements are of finite order.

EDIT: This actually isn't true - yes if such elements formed subgroups those subgroups would automatically be normal. However, the product of two torsion elements is not necessarily torsion (see anything on Burnside's Problem). If, however, the set of non-torsion elements was a subgroup my statement would be saved (as then the intersection would contain elements of infinite order, but we have proven that they are all of finite order). So...does the product of two elements of infinite order product an element of infinite order?

**poorna** · August 14th 2009, 09:25 PM

@ TheAbstractionist

Why do you say x is of even order. ie why is it that x is an element of <x^2>

**TheAbstractionist** · August 16th 2009, 08:09 AM

I DID NOT say that $x$ is of even order.

$x$ is a nontrivial element belonging to every nontrivial subgroup of $G.$ If $x^2\ne1_G,$ then $\langle x^2\rangle$ is a nontrivial subgroup of $G$ and so contains $x.$

**aman_cc** · August 23rd 2009, 07:53 AM

Am I correct in saying that they can't have any element (except 'e') common.

Reason: If o(a) = inifinity (as is assumed) => a^i != a^j for different i,j.

So let a^i be common to all such sub-groups. Consider the sub-group generated by a^(i+1). Obviously a^i doesn't belong to this sub-group. Hence contradiction.

Please let me know if my argument is correct and appropriate.

Originally Posted by Opalg

Suppose that G has an element a of infinite order. What can you then say about the sequence of subgroups generated by $a^2,\ a^3,\ a^4,\ldots\,?$

**alunw** · August 23rd 2009, 12:51 PM

Originally Posted by Swlabr

So...does the product of two elements of infinite order product an element of infinite order?

Are you asking if the product of two non-torsion elements of a group is necessarily a non-torsion element?
The answer to this is certainly not. It is easy to find 2 elements of SL(2,C) (2*2 complex matrices with determinant 1) with any desired traces (and if the trace is not real and between -2 and 2 the matrix has infinite order) such that the product of the two matrices has any other desired trace (so you can arrange for it to be one of the values that means the matrix has finite order). In fact you can do much more complicated things like finding two matrices A,B whose commutator (A*B*A^-1*B^-1) has a specified finite order, and where two designated "special words" (such as A^5*B and A*B^2) also have possibly different finite orders.
Discrete subgroups of SL(2,C) (or more properly PSL(2,C)) are called Kleinian groups, and have been very extensively studied in recent years because they can be used to represent hyperbolic isometries and hence are important to hyperbolic manifold theory. They are often associated with very interesting fractals (the "limit set" of the group). Even in the case of 2 generator groups you can arrange for the group to have several different kinds of torsion element.
I posted an album with a couple of pictures of Kleinian limit sets.

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