# Free Products - representation of 1 by empty word is unique

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.

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 $$(0,0)$$.

• 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. Commented Apr 14, 2021 at 16:13

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.

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