0
$\begingroup$

I am studying Free Products of Groups from Munkres' Topology book. The definition of free products is:

Let $G$ group, $\{G_a\}_{a \in J}$ family of groups that generates $G$ and $G_a \cap G_b$ consists of the idenity element alone whenever $a \neq b$. We say that $G$ is the free product of $G_a$, if for each $x \in G$ there is only one reduced word that represents $x$.

After that it says and proves, that it suffices to know that the representation of $1$ by the empty word is unique.

What does exactly the latter means? Isn't the empty word unique and doesn't always represents $1$? By the previous argument it looks to me like the condition of the uniqueness of the reduced word to be redundant.

$\endgroup$

2 Answers 2

1
$\begingroup$

Munkres defines a word as finite sequence $s = (x_1,\dots,x_n)$ of elements $x_i \in G_{\alpha_i}$ and says that such a sequence represents $x \in G$ if $x = \prod_{i=1}^n x_i = x_1\dots x_n$. As an adhoc notation let us write $\prod s = \prod_{i=1}^n x_i$. Then he defines two reduction operations shortening sequences $s$ of length $n$ to sequences $s'$ of length $n-1$ such that $\prod s = \prod s'$. One of these operations is the omission of $x_i$ if $x_i = 1$.

At the beginning he does not say anything about $n$ and thus one might think that we only deal with sequences of length $n \ge 1$. However, after he defined the reduction operations he explicitly makes the convention that the sequence $\emptyset$ of length $0$ (i.e. the empty word) is regarded as a reduced word which represents $1$. This is common practice; for example in elementary calculus one frequently defines the empty sum as $0$ and the empty product as $1$ (i.e. as the neutral elements with respect to additon and multiplication).

Let us come to your question. The empty word always represents $1$. But there may be other reduced words $s$ representing $1$ and then the representation of $1$ as a reduced word is not unique. This is what Munkres means. But admittedly the formulation "the representation of $1$ by the empty word is unique" may be misleading. He should better have said "the representation of $1$ by a reduced word is unique, and this is the representation by the empty word".

Here is an example. Let $G = \mathbb Z \oplus \mathbb Z$. Let $G_1$ be the subgroup of all pairs $(a,0)$ and $G_2$ the subgroup of all pairs $(0,b)$. Their intersection only contains the additive identity element $(0,0)$. Clearly each element $(a,b)$ of $G$ is represented by the word $((a,0),(0,b))$. But all words $((a,0),(0,b),(-a,0),(0,-b))$ with $a,b \ne 0$ are reduced and represent $(0,0)$.

$\endgroup$
1
  • $\begingroup$ Thank you very much, actually that's what i was searching for. Firstly i thought the same on my own. But really the way he describes it got me confused. I thought for a while that empty word isn't unique. $\endgroup$
    – Nash-iOS
    Commented Apr 14, 2021 at 16:13
0
$\begingroup$

Let me reword that sufficient condition in order to make it more explicit:

... it suffices to know that the representation of $1$ by the empty word is the unique representation of $1$ by a reduced word.

$\endgroup$
2
  • $\begingroup$ Thanks for the respond, if you add more words wouldn't that make the word not empty (or not reduced) ? $\endgroup$
    – Nash-iOS
    Commented Apr 14, 2021 at 13:57
  • $\begingroup$ I fear you misunderstood my "adding more words" comment, so I have rewritten that part. $\endgroup$
    – Lee Mosher
    Commented Apr 14, 2021 at 14:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .