Amalgamated Product example.

Question

I have been reading about amalgamated products online where they are defined as the quotient group formed by the free product and a normal subgroup of the free product.

I am having a hard time understanding the following example we were given in class,

$$ \mathbb{Z}_2 *_{e} \mathbb{Z}_2 = < a,b \mid a^2=e, \hspace{4mm} b^2=e>$$

First, I am not sure how to write the free product of $\mathbb{Z}_2 * \mathbb{Z}_2$, both groups are generated by $1$ so I am assuming the free product would just be $<1>$. However, this can't be right because it looks like the fact that $\mathbb{Z}_2$ is generated by $1$ is not taken in to consideration and just left arbitrary, by writing the generators as $[a]$ and $[b]$. Why is this happening? Why are we writing that $[a]$ generates $\mathbb{Z}_2$ when we already know that $1$ does??

I think I don't understand the free product completely.

Answer 1

Here's how I think about the free product. Consider the free product of two arbitrary groups $G$ and $H$, i.e. $G\ast H$. Elements of this group looks like $$ g_1h_1g_2h_2\cdots g_nh_n,\quad g_i\in G,h_j\in H. $$ We get a string of elements from $G$ and $H$ in a row. Usually we call the group elements letter and the string of elements a word. Multiplication is done by concatenation, e.g. $$ gh\cdot g'h'=ghg'h'. $$ When we ever have the letter $e_G$ or $e_H$ in the sequence, we throw it out. For example, $$ gh\cdot e_He_G=ghe_G=gh. $$ Basically, any word you write consisting of only the identity elements we associate to the empty word $\varepsilon$. This is the word that, when you concatenate it onto any other word, adds no letter. Hopefully this is enough intuition to get you started.

In the case of $\mathbb Z_2\ast\mathbb Z_2$, we want to consider strings of elements of each copy of $\mathbb Z_2$. As you note, they're both generated by $1$. We can't just write $$ 1111111111111111 $$ for group elements though. We have to distinguish them somehow, and that's where we call one of them $a$ and the other one $b$. The words are then of the form $$ ababababababab\cdots\text{ or }babababa\cdots $$

Answer 2

When you take the free product of two groups, you think of the groups as being made from different elements. The names $a$ and $b$ represent the element 1 in the first and second copies of $\mathbb{Z}_2$, respectively. They had to use the arbitrary names $a,b$ to distinguish the element 1 in each copy of $\mathbb{Z}_2$.

The idea being the free product is to have a general way to create a group $G*H$ which contains two given groups $G$ and $H$. This is accomplished by letting the elements of $G*H$ be words whose letters are elements of $G$ or $H$, where replacing two adjacent letters from the same group with their product in that group doesn't is said to result in the same word.

Answer 3

First of all, the two 1s are not the same … moreover, I assume you are saying they are generated by $1$ when looking at the operation as the usual sum of integers. Dealing with a product (free or amalgamated), the addition notation for the group operation is not the wisest choice.

First step, presentation of a group. $G=\langle a\ \vert\ a^2=e\rangle$ is a group which has one generator (let's call it $a$) and where one relation holds (namely $a^2=e$). The generators are specified in the first part of the writing, the relations in the second. Relations are simply expressions involving the generators (with product notation for the operation in the group) which are usually equal to the identity of the group. The group that I have just presented is $\mathbb{Z}_2$: you have only one generator, and when you multiply it by itself (remember: product notation!) it gives you identity, so the group will be $G=\{e,a\}$ with product rule $ea=ae=a$, $e^2=a^2=e$. No matter if you call them $0,1$ or $a,b$ or Joe and Frank. The group is always the same.

Now, the free product. Free product means just words made up by elements of the two groups, where you can modify the word only if there are two elements of the same group one aside the other, then you can substitute to them their product.

For example, take $\mathbb{Z}_3=\langle a\ vert\ a^3=e\rangle$ and $\mathbb{Z}=\langle b\rangle$. So $$\mathbb{Z}\ast \mathbb{Z}_3=\langle a, b\vert a^3=e\rangle$$ i.e. all the words made by $a$'s and $b$'s where the only simplifications you are allowed to make are those equivalent to $a^3=e$. So $ab^7ab^6$ is equivalent to $a^4b^7ab^6$ or $a^4b^7a^7b^6$ or $a^{-2}b^7ab^6a^3$ and so on…

Lastly, an amalgamated product is a free product where you have thrown in some other relations. These additional relations must come from some identification between elements of one group and elements of the other. It is easy to see that you always need to identify elements of two subgroups, which can be in some way related. So, you pick $H$ such that there are maps $f_1:H\to G_1$ and $f_2:H\to G_2$; now you want to construct an "almost" free product, where these two images of $H$ are the same. What do you do? Well, you just impose all the relations $f_1(h)=f_2(h)$ for $h\in H$ when you construct the presentation. If you were kind enough to take $H$ as a finitely generated group, you'll have generators $h_1,\ldots, h_k$, so, if $$G_1=\langle a_1,\ldots, a_n\ \vert \ R_1\rangle$$ $$G_2=\langle b_1,\ldots, b_m\ \vert\ R_2\rangle$$ where $R_1$, $R_2$ are sets of relations, then $$G_1\ast_H G_2=\langle a_1,\ldots, a_n, b_1,\ldots, b_m\ \vert\ R_1\cup R_2\cup\{f_1(h_j)=f_2(h_j)\ j=1,\ldots, k\}\rangle$$ This means you can also make simplifications among elements of $G_1$ and $G_2$ whenever they belong to the images of $H$.

In your example, $H=\{e\}$, so you are not allowing any further simplification, hence your amalgamated product is just the free product, where you denoted by $a$ the generator of one group and $b$ the generator of the other, in order to distinguish them (you could have written something like $1_1$ and $1_2$ but it would have been awkward…).