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At the other end, I think that a word-hyperbolic group $G$ has a uniformity defined by Gromov whose completion points are called the Gromov boundary. However, multiplication is not uniformly continuous in this construction; rather you only have that the uniformity is left-invariant. Thus, the completion is not a topological group, but it does have a left action of $G$. You can only define right-infinite products, and you can only multiply them by group elements on the left. Still, you can do that much, which still lets you consider many interesting infinite relations.
Your example of groups of matrices is modeled by a uniformity of intermediate quality. The uniformity of the additive Lie group structure on $M_3(\mathbb{C})$ is not the same as the uniformity of the multiplicative Lie group structure on $GL_3(\mathbb{C})$, even though the topologies are the same. Multiplication is uniformly continuous in extends to the formercompletion, but the group inverse law is inversion does not. (Toy model: The function $1/x$ on $\mathbb{R}^\times$ is not uniformly continuous in the standard metric on $\mathbb{R}$.) \mathbb{R}$and obviously does not extend to$0$.) The completion of$GL_3(\mathbb{C})$is obviously$M_3(\mathbb{C})$, and any subgroup has an inherited uniformity and a completion which is then a semigroup. So an answer to your second question is that a discrete group can have many different semigroup semigroup-completed uniformities, some of which come from matrix representations. 1 This answer is partly an expansion of the comments so far. What you are looking for is a type of decoration on groups that (1) lets you evaluate infinite products, and (2) also creates extra points that are the value of some or all of those products. There is one fairly inevitable answer to (1): You should consider topological groups or at least topological spaces, since taking limits in a topology is the main way to extend finite products to infinite products. But for (2), a topology isn't good enough. You get more information from a metric than from a topology, because a metric space has a completion. On the other hand, a metric isn't entirely satisfactory either, among other reasons because it has extra, non-canonical data. A good alternative to a metric is a uniformity. A uniform space also has a completion (in which all Cauchy nets converge) and every metric space is canonically a uniform space. In fact, every well-behaved topological group has a canonical associated uniformity, because you can use the neighborhoods of the identity to make uniform neighborhoods. (I am a little hazy on how exactly the topological group should be well-behaved. It should be sufficient for it to be Hausdorff and for the left and right uniformities to agree.) Thus every such topological group has a completion which is also a topological group, and which then has infinite products which are group elements. As Mark Sapir suggests, one such uniformity is the profinite uniformity on a residually finite group. At the other end, I think that a word-hyperbolic group$G$has a uniformity defined by Gromov whose completion points are called the Gromov boundary. However, multiplication is not uniformly continuous in this construction; rather you only have that the uniformity is left-invariant. Thus, the completion is not a topological group, but it does have a left action of$G$. You can only define right-infinite products, and you can only multiply them by group elements on the left. Still, you can do that much, which still lets you consider many interesting infinite relations. Your example of groups of matrices is modeled by a uniformity of intermediate quality. The uniformity of the additive Lie group structure on$M_3(\mathbb{C})$is not the same as the uniformity of the multiplicative Lie group structure on$GL_3(\mathbb{C})$, even though the topologies are the same. Multiplication is uniformly continuous in the former, but the group inverse law is not. (Toy model: The function$1/x$on$\mathbb{R}^\times$is not uniformly continuous in the standard metric on$\mathbb{R}$.) The completion of$GL_3(\mathbb{C})$is obviously$M_3(\mathbb{C})\$, and any subgroup has an inherited uniformity and a completion which is then a semigroup. So an answer to your second question is that a discrete group can have many different semigroup uniformities, some of which come from matrix representations.