- #1
Oxymoron
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It seems that I am having difficulty understanding some of the definitions given to me concerning Free Groups.
I just finished a thread on free groups where I learned that a free group is one that is constructed from an underlying basis set that is enlarged so that one may form all finite words - then declare that two words are equivalent if and only if inverses and identity cancellation would make them so.
Then I discovered that all free groups have a universal mapping property - that is, any function from the basis set S into a group G has a unique extension to a homomorphism of the free group F(S) into G.
I've now come across the concept of a free product and as far as I can tell we adopt a similar approach to their construction as we do for free groups. Let me explain...
First, let S be a collection of groups
[tex]S = \{G_1,G_2,\dots\}[/tex]
as opposed to a set of generator elements as for free groups. Then we define the free product
[tex]\star_{i\in I} G_i = G_1\ast G_2\ast \dots[/tex]
as the most general group containing all the [itex]G_i[/itex]'s as subgroups. And likewise, the free product satisfies a universal mapping theorem: Given any collection of homomorphisms from each factor group into some common group W, there is a unique extension to a homomorphism of the free product into W.
Now I am not going to pretend that I am an expert at any of this, in fact, quite the opposite. I want to be able to "prove that [itex](G_1 \ast_A G_2) \ast_A G_3[/itex] is isomorphic to [itex]G_1 \ast_A (G_2 \ast_A G_3)[/itex] where A is a subgroup of the three groups [itex]G_i[/itex] and where amalgams are formed with respect to the respective inclusions of the group A are isomorphic."
In order to do so I will need a lot of help to understand:
What does it mean to be isomorphic in this sense?
What on Earth is an amalgam?
Is the free product of two groups A and B necessarily free? Is the free product free if and only if A and B are free?
I know this is a lot of stuff to do, but if anyone can help flatten out the learning curve with this, I will rest easy.
Thanks.
I just finished a thread on free groups where I learned that a free group is one that is constructed from an underlying basis set that is enlarged so that one may form all finite words - then declare that two words are equivalent if and only if inverses and identity cancellation would make them so.
Then I discovered that all free groups have a universal mapping property - that is, any function from the basis set S into a group G has a unique extension to a homomorphism of the free group F(S) into G.
I've now come across the concept of a free product and as far as I can tell we adopt a similar approach to their construction as we do for free groups. Let me explain...
First, let S be a collection of groups
[tex]S = \{G_1,G_2,\dots\}[/tex]
as opposed to a set of generator elements as for free groups. Then we define the free product
[tex]\star_{i\in I} G_i = G_1\ast G_2\ast \dots[/tex]
as the most general group containing all the [itex]G_i[/itex]'s as subgroups. And likewise, the free product satisfies a universal mapping theorem: Given any collection of homomorphisms from each factor group into some common group W, there is a unique extension to a homomorphism of the free product into W.
Now I am not going to pretend that I am an expert at any of this, in fact, quite the opposite. I want to be able to "prove that [itex](G_1 \ast_A G_2) \ast_A G_3[/itex] is isomorphic to [itex]G_1 \ast_A (G_2 \ast_A G_3)[/itex] where A is a subgroup of the three groups [itex]G_i[/itex] and where amalgams are formed with respect to the respective inclusions of the group A are isomorphic."
In order to do so I will need a lot of help to understand:
What does it mean to be isomorphic in this sense?
What on Earth is an amalgam?
Is the free product of two groups A and B necessarily free? Is the free product free if and only if A and B are free?
I know this is a lot of stuff to do, but if anyone can help flatten out the learning curve with this, I will rest easy.
Thanks.