# Limit of a sequence of algebraic structures

(Disclaimer: I'm a novice in algebra. I'm at Chapter 13 of this free online abstract algebra book. This is a , because it's a loose "does this idea make sense" kind of question, instead of "solve this problem".)

I've been thinking about sequences of groups that, in some sense, appear to get closer and closer to some infinite group that has all of the previous groups as subgroups.

For example, the sequence

$$\mathbb Z_2,~~~ \mathbb Z_2 \times \mathbb Z_2,~~~ \mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2,~~~ \dots$$

in a certain sense approaches the group $\mathbb Z_2^\infty$ (I'm not sure if it has a more common symbol) of infinite binary strings under the bitwise XOR operation.

Similarly, the sequence

$$S_1,~ S_2,~ S_3,~ \dots$$

approaches $S_\mathbb N$, the symmetric group on all natural numbers. So you could be silly and write: $\lim_{n \to \infty} S_n = S_\mathbb N$.

Does the pattern I'm seeing make any sense? I'm interested to see if these "sorta-limits" can be given a more rigorous definition. Is there some way to define the "distance" between two groups, so that the usual definition1 for sequences can apply to them, and do the above patterns work out under that definition?

Of course, there's no reason to restrict the question to groups: does this make sense for sets, monoids, rings, ...?

[1] i.e., the sequence gets closer and closer to some limit group $L$, and to no other group.

• They are called limits and colimits (or sometimes inverse limits and direct limits). See here. – Tunococ Jan 9 '16 at 15:27
• I have vaguely heard about those terms, but I know next to nothing about category theory. (I basically know what objects, morphisms, functors, and natural transformations are.) If they answer my question, I'd like a simple explanation how. (I'll take a more thorough look later, but at a glance, I can barely understand the Wikipedia article you linked, and it doesn't even mention the word sequence. :() – Lynn Jan 9 '16 at 15:31
• See these two pages then: direct limits and inverse limits. A sequence is a special case of a direct system. (Trust me it is worth learning the category-theoretic vocabulary. It is not conceptually difficult.) – Tunococ Jan 9 '16 at 15:43
• You have to be a bit careful: If you consider $S_1 \hookrightarrow S_2 \hookrightarrow S_3 \hookrightarrow \dotsb$ then the (co)limit consists of all permutations of $S_\mathbb{N}$ that fix all but finitely many elements. So we only get a rather small subgroup of $S_\mathbb{N}$. – Jendrik Stelzner Jan 9 '16 at 16:21
• @JendrikStelzner: Ah, that does make sense. Thanks. – Lynn Jan 9 '16 at 19:39

I think the easiest case to understand is when you have an inclusion of groups $$R_1 \subseteq R_2 \subseteq \cdots$$ which are all subgroups of a very large group $\mathcal R$. Now the natural "limit" of this progression of groups is the union $$R = \bigcup\limits_{i=1}^{\infty} R_i$$ Usually, the union of a collection of subgroups of a given group is not a group, but here $R$ is a group. Also, $R$ has a special property which you can immediately verify: (1) If $S$ is any other subgroup of $\mathcal R$, and all the $R_i$ are contained in $S$, then $R$ is contained in $S$.

This property (1) is pretty obvious, but it admits a useful generalization (also easy to prove): (2) If $S$ is any group whatsoever, and $f_i: R_i \rightarrow S$ is a collection of group homomorphisms such that the restriction of $f_i$ to $R_{i-1}$ is $f_{i-1}$, then there is a unique homomorphism $f: R \rightarrow S$ such that $f_{|R_i} = f_i$ for all $i$.

The reason why (2) is useful is you might have a more general situation. Instead of having an inclusion of groups $R_1 \subseteq R_2 \subseteq \cdots$, you might instead have a sequence of injective group homomorphisms $$R_1 \xrightarrow{h_1} R_2 \xrightarrow{h_2} R_3 \cdots$$

This is basically the same situation as before, since each $R_i$ can be thought of as a subgroup of $R_{i+1}$, but transitivity of subgroup inclusions can only be done finitely many steps at a time. You may wish to obtain a similar "limit" of the groups $R_1, R_2, ...$, but this time you cannot really take "the union" of the groups $R_1, R_2, ...$, because logically it only makes sense to take a union of sets which are all contained in a common larger set.

A fantastic exercise to aid your understanding of "direct limits" (of which everything I've said in this answer is a special case) is to explicitly construct a group $R$ which is an analogue of the "union" of these groups $R_1, R_2$ etc. under the group homomorphisms $R_1 \xrightarrow{h_1} R_2 \xrightarrow{h_2} \cdots$. You will know you have the right construction when you can embed the groups $R_i$ into $R$, and show that a similar property to (2) holds. When I was an undergrad, I encountered very similar concepts and was naturally led to thinking about problems like this. I formulated and solved the exact problem I suggested to you, which took me about 6 hours on a single afternoon (I'm pretty slow, it probably won't take you that long) and only afterwards learned that this was a special case of a more general notion of a "direct limit."

I am not sure whether $S_{\mathbb{N}}$ is really a limit in the same sense as I'm talking about (the natural thing to do would be to give a group monomorphism $h_n: S_n \rightarrow S_{n+1}$, where $h_n(\sigma)$ always fixes the element $(n+1)$, but I think if you constructed the "limit" as I have described it (that is, you took the "union" of all the $S_n$), you would end up with a subgroup of $S_{\mathbb{N}}$ consisting of those permutations which fixed all but finitely many of the elements.

As for $Z_2 \times Z_2 \times \cdots$ being a "limit" of the groups $Z_2, Z_2 \times Z_2, ...$ that is a slightly different kind of limit than the one I described. Let $R = Z_2 \times Z_2 \times \cdots$, and let $R_n = \prod\limits_{i=1}^n Z_2$. Here, instead of having injective homomorphisms $R_1 \rightarrow R_2 \rightarrow \cdots$, you will actually want to work with surjective homomorphisms $$\cdots \rightarrow R_3 \rightarrow R_2 \rightarrow R_1$$ Basically, $R$ is still a limit of the $R_n$, but in a "dual sense."

• Thank you very much for this answer! In your exercise, you write that $R_1, R_2$ etc. are groups, but $h_1$, $h_2$ are ring homomorphisms; did you mean for the $R_i$ to be rings? – Lynn Jan 9 '16 at 19:37
• I originally wrote this for rings, but then I realized all your examples were groups, so I changed it to groups. Everything I said applies equally well for groups or rings when it comes to limits. – D_S Jan 9 '16 at 20:31