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 ? Am I doing this right?
Thanks!
The identity element of is or , a coset. This coset corresponds precisely to the subgroup (the elements are the same).
Now, if has finite order then there exists an such that . Thus, .
(The bit where you go ` ' holds because )
In multiplicative notation, if has finite order then there exists an such that . Thus, .
(The bit where you go ` ' holds because )
What do you not understand? Bruno's comment, or the proof?
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 then ).
He also pointed out that in an abelian group, this set S will always be a group, as where .
Do not worry about what a free-product is. You will learn that with time.