# Groups - Quotient groups

• Aug 12th 2010, 02:54 AM
Groups - Quotient groups
Let G be a group, A be a group of all elements in G that have a finite order.

Prove that the quotient group G/A contains no elements of finite order.

Hmmm... If I assume that there is a gA in G/A for which there exists a natural n so that |gA|=n , then I get that : gA+...gA=gA (n times)

How do I continue (Thinking)? Am I doing this right?

Thanks! :)
• Aug 12th 2010, 03:54 AM
Swlabr
Quote:

Let G be a group, A be a group of all elements in G that have a finite order.

Prove that the quotient group G/A contains no elements of finite order.

Hmmm... If I assume that there is a gA in G/A for which there exists a natural n so that |gA|=n , then I get that : gA+...gA=gA (n times)

How do I continue (Thinking)? Am I doing this right?

Thanks! :)

If $n(gA)=0A$ (i.e. $gA$ has order $n$) then this means that $ng \in A$ (why?). The contradiction now follows.

(To write it in non-abelian notation, 'If $(gH)^n=1H$ (i.e. $gA$ has order $n$) then this means that $g^n \in H$')
• Aug 12th 2010, 04:21 AM
I see... although I don't really understand why ng is in A - is it because : (ng)A=(e)A=A, or is it because A is the neutral in G/A (Thinking)

It's a little complicated... Literally, Abstract Algebra...
• Aug 12th 2010, 04:31 AM
Swlabr
Quote:

I see... although I don't really understand why ng is in A - is it because : (ng)A=(e)A=A, or is it because A is the neutral in G/A (Thinking)

It's a little complicated... Literally, Abstract Algebra...

The identity element of $G/A$ is $eA$ or $0+A$, a coset. This coset corresponds precisely to the subgroup $A$ (the elements are the same).

Now, if $g+A$ has finite order then there exists an $n$ such that $n(g+A) = ng+A = 0+A$. Thus, $(ng)-0 \in A \Rightarrow ng \in A$.

(The bit where you go  $(ng)-0 \in A$' holds because $a+A = b+A \Leftrightarrow a-b \in A$)

In multiplicative notation, if $gH$ has finite order then there exists an $n$ such that $(gH)^n = (g^n)H = 1H$. Thus, $g^n1^{-1} \in H \Rightarrow g^n \in H$.

(The bit where you go  $g^n1^{-1} \in H$' holds because $gH = hH \Leftrightarrow gh^{-1} \in H$)
• Aug 12th 2010, 05:22 AM
I see... Thank you :)

So the contradiction is a consequence of the idea that ng is in A?

I mean, if ng is in A, then there exists an m for which (ng)^m=1, means that mng=1, but I don't see the contradiction here :O
• Aug 12th 2010, 05:44 AM
Swlabr
Quote:

I see... Thank you :)

So the contradiction is a consequence of the idea that ng is in A?

I mean, if ng is in A, then there exists an m for which (ng)^m=1, means that mng=1, but I don't see the contradiction here :O

What has just been proven is that if $g+A$ has finite order then $g \in A$. Therefore, $g+A=0+A$. Apply this to what you are trying to prove.
• Aug 12th 2010, 05:59 AM
Hmmm.. I still can't see the contradiction. Besides, how did you get to g+A, while we were discussing gn+A?
• Aug 12th 2010, 06:07 AM
Swlabr
Quote:

Hmmm.. I still can't see the contradiction. Besides, how did you get to g+A, while we were discussing gn+A?

If $g+A$ has finite order, $n$, then $gn \in A$ and so $gn$ has finite order (by the definition of $A$) and so there exists an $m$ such that $g(mn)=0$. Thus, $g$ has order $mn$, which is finite. Therefore, $g \in A$ by the definition of $A$...and so $g+A=0+A$.
• Aug 12th 2010, 06:17 AM
I guess I'm a having some trouble figuring this out :P
I understand why what you've just said is correct, except that I don't understand why g+A=0+A, or how this leads to a contradiction.

:)
• Aug 12th 2010, 06:19 AM
Swlabr
Quote:

I guess I'm a having some trouble figuring this out :P
I understand why what you've just said is correct, except that I don't understand why g+A=0+A, or how this leads to a contradiction.

:)

g+A=0+A as g is contained in A. It is contained in A because g must have finite order, and A is the group of all elements of finite order.
• Aug 12th 2010, 07:22 AM
Quote:

Originally Posted by Swlabr
g+A=0+A as g is contained in A. It is contained in A because g must have finite order, and A is the group of all elements of finite order.

Fine. Then everything is perfect. Where is the contradiction :D?
• Aug 12th 2010, 07:30 AM
Swlabr
Quote:

Fine. Then everything is perfect. Where is the contradiction :D?

Emm...I was assuming $g+A \neq 0+A$, but I suppose you don't have to.
• Aug 12th 2010, 10:19 AM
Bruno J.
Quote:

Let G be a group, A be a group of all elements in G that have a finite order.

Prove that the quotient group G/A contains no elements of finite order.

Hmmm... If I assume that there is a gA in G/A for which there exists a natural n so that |gA|=n , then I get that : gA+...gA=gA (n times)

How do I continue (Thinking)? Am I doing this right?

Thanks! :)

I'd just like to remark that the set of all elements of $G$ which have finite order is not in general a subgroup of $G$; for example, take the free product of two finite groups.

In an abelian group, however, it is always a subgroup.
• Aug 12th 2010, 10:30 PM
Let S be the set of all elements of finite order in a group. Then it is possible to multiply two elements together to get an element of infinite order (for example, you can find 2x2 matrices which are of finite order multiplicatively, but when multiplied together they will form a matrix of infinite multiplicative order). This means that S is not necessarily a subgroup of your group. However, if S is a group, it will be normal, as conjugation preserves the order of an element (if $g^n=1$ then $(h^{-1}gh)^n = h^{-1}g^nh=1$).
He also pointed out that in an abelian group, this set S will always be a group, as $n(a+b)=1$ where $n= o(a)\cdoto(b)$.