# How do I construct this quasicyclic group?

Question:

Let $p$ be a prime number. Let $G_n=C_{p^n}$ be the cyclic group of order $p^n$ with generator $x_n$. We define $\varphi:G_n \rightarrow G_{n+1}$ by $\varphi(x_n^a)=x_{n+1}^{pa}$. Using the above construction we obtain a group $G$, called a quasicyclic group and usually denoted $C_{p^\infty}$. Prove that $C_{p^\infty} \cong H$ where $H=\{ z \in \mathbb{C}^\times: \exists n \: z^{p^n}=1 \}$.

The lecturer didn't include the "above construction" and I can't find any more detail on the net. Anyone know how to construct this group from the homomorphisms given?

-
@Qiaochu: I'd found that web page and didn't really understand what it was saying. @Lubos: I don't think I made my question clear enough. I'd already found that isomorphism (between $C_{p^n}$ and $H_n$) and made the observation given in your last paragraph. What I need help with is the limit bit, which I still don't understand. I don't know anything about this "limit" idea, not even enough for me to see how it's valid that two sequences of isomorphic groups have isomorphic limits. This is what I'd like explaining, please! – Kate May 26 '11 at 16:55
Dear Kate, if two sequences are isomorphic one element after another, then all properties of these two sequences including the limit have to be isomorphic as well. "Isomorphic" really means "the same thing when it comes to any behavior of them". – Luboš Motl May 26 '11 at 18:33

Like Qiaochu says, and Luboš explains, this is a directed limit of groups.

We don't need the full machinery of direct limits, because the index set is a chain instead of merely a directed set, so I'll simplify the description of the construction to that instance. Nonetheless, some of the arguments may seem "needlessly complicated" given that what you have are embeddings; I make them that way so that they generalize more easily to more general situations.

Suppose that for every positive integer $n$ we have a group $G_n$, and for any $m\leq n$ we have a homomorphism $f_{mn}\colon G_m\to G_n$ (note the order) that satisfy the following properties:

1. $f_{nn}=\mathrm{id}$ for every $n$; and
2. If $m\leq n\leq p$, then $f_{mp} = f_{np}\circ f_{mn}$.

You should think of this as an "infinite sequence" of groups that extends off to infinity on the right: $$G_1 \longrightarrow G_{2} \longrightarrow G_{3} \longrightarrow \cdots \longrightarrow G_{k-1} \longrightarrow G_k \longrightarrow G_{k+1} \longrightarrow \cdots$$ (we really have more maps, like a map from $G_1$ directly to $G_4$, but condition 2 above says that the map that goes directly from $G_1$ to $G_4$ is the same as the map we get by going from $G_1$ to $G_2$ to $G_3$ and then to $G_4$, so we can just focus on the maps from $G_{n}$ to $G_{n+1}$).

Now, it makes no sense in general to ask how much is the product of, say, an element in $G_2$ times an element of $G_{10}$: they live in different groups, so there's no multiplication defined. But we can take $g\in G_2$, map it to $G_{10}$ using $f_{2,10}$, and then multiply it by the element of $G_{10}$, and this makes sense. In fact, given any two elements in the disjoint union of the $G_n$, we can map the two elements to the same group and multiply them then. Of course, we can also map them to a larger group, and we'll get something else (since it is in a different group), which is a bit of a problem...

So we want to take all of this information and put it together into a coherent whole that will give a well-defined multiplication.

One way to do this is the following: we will first approximate things by working on the disjoint union of the $G_i$; to make sure we know where each element "lives", we will "tag" each element with the group it comes from. That is, our (first approximation to an) underlying set will consist of elements of the form $(g,G_n)$, where $g\in G_n$.

How do we define multiplication? Well, if we have $(g,G_n)$ and $(h,G_m)$, and $n=m$, then it's easy: just multiply: $(g,G_n)\cdot(h,G_n) = (gh,G_n)$.

If $n\neq m$, then we want to follow the idea of "map them both to the same group and multiply them there." So let $k=\max\{n,m\}$, and we define $$(g,G_n)\cdot (h,G_m) = \bigl( f_{nk}(g)f_{mk}(h), G_k\bigr).$$

Unfortunately, this does not define a group structure on the disjoint union. For one thing, note that $(e,G_n)$ works as an identity for every element of the form $(g,G_n)$, but so does $(e,G_k)$ for any $k\lt n$. That is, we may have $\mathbf{x}\mathbf{y}=\mathbf{x}\mathbf{z}$, but $\mathbf{y}\neq\mathbf{z}$, with $\mathbf{x}$, $\mathbf{y}$, and $\mathbf{z}$ in the disjoint union. This is certainly not a group, then.

In order to make sure it's a group, then we need certain elements of the disjoint union to be "the same": for example, if $n\lt m$, then $(g,G_n)\cdot(h,G_{m})$ is going to give the same answer as $(f_{nk}(g),G_k)\cdot(h,G_m)$ for every $k$, $n\leq k\leq m$. So all of those have to be "equal."

So we define an equivalence relation on the set of pairs $(g,G_n)$ as follows: $(g,G_n)$ and $(h,G_m)$ should be "the same" if they eventually map to the same thing. That is: $$(g,G_n)\sim (h,G_m)\Longleftrightarrow \text{there exists }k\geq\max\{m,n\}\text{ such that }f_{nk}(g)=f_{mk}(h).$$ The conditions of the maps $f_{ij}$ ensure that this relation i reflexive and transitive (it is plainly symmetric). So our underlying set will be the set of equivalent classes under $\sim$, the set of all $[(g,G_n)]$, where $g\in G_n$, and $[(g,G_n)] = [(h,G_m)]$ if and only if there exists $k\geq \max\{n,m\}$ such that $f_{nk}(g)=f_{mk}(h)$.

Then we define multiplication of classes as follows: given $[(g,G_n)]$ and $[(h,G_m)]$, let $k$ be any integer greater than or equal to both $n$ and $m$, and define $$[(g,G_n)]\cdot[(h,G_m)] = \bigl[(f_{nk}(g) f_{mk}(h), G_k)\bigr].$$

Proposition. The product operation is well-defined.

Proof. First, we show that the choice of $k$ does not matter: suppose $k,\ell\geq\max\{m,n\}$. Let $p\geq\max\{k,\ell\}$. Then \begin{align*} f_{kp}\bigl(f_{nk}(g)f_{mk}(h)\bigr) &= f_{kp}f_{nk}(g)\cdot f_{kp}f_{mk}(h)\\ &= f_{pn}(g)f_{pm}(h)\\ &= f_{p\ell}\circ f_{n\ell}(g)\cdot f_{p\ell}\circ f_{m\ell}(h)\\ &= f_{p\ell}(f_{n\ell}(g)f_{m\ell}(h)). \end{align*} Therefore, $[(f_{nk}(g)f_{mk}(h),G_k)] = [(f_{pn}(g)f_{pm}(h),G_p)] = [(f_{n\ell}(g)f_{m\ell}(h), G_{\ell})]$. So the choice of $k$ does not matter, assuming we leave the representatives fixed.

Next, suppose that $[(g,G_n)] = [(a,G_p)]$ and $[(h,G_m)]=[(b,G_q)]$; we want to show that if $k\geq \max\{n,m\}$ and $\ell\geq\max\{m,q\}$, then $[(f_{nk}(g)f_{mk}(b),G_k)] = [(f_{p\ell}(a)f_{q\ell}(b),G_{p})]$.

We know that there exists $i\geq n,p$ such that $f_{ni}(g)=f_{pi}(a)$, and there exists $j\geq m,q$ such that $f_{mj}(h)=f_{qj}(b)$. Let $t=\max\{i,j\}$. Then \begin{align*} f_{nt}(g) &= f_{it}\circ f_{ni}(g)\\ &= f_{it}\circ f_{pi}(a)\\ &= f_{pt}(a).\\ f_{mt}(h) &= f_{jt}\circ f_{mj}(h)\\ &= f_{jt}\circ f_{qj}(b)\\ &= f_{qt}(b). \end{align*} Therefore, $f_{nt}(g)f_{mt}(h) = f_{pt}(a)f_{qt}(b)$, so $$[ (g,G_n)]\cdot[(h,G_m)] = [(f_{nt}(g)f_{mt}(h),G_t)] = [(f_{pt}(a)f_{qt}(b),G_t)] = [(a,G_p)]\cdot[(b,G_q)],$$ as desired. QED

Theorem. The set of equivalence classes $[(g,G_n)]$ with the operation $\cdot$ is a group.

The only annoying part of verifying the above is showing associativity; the identity element is given by the class of $(e,G_n)$ for any $n$, and the inverse of $[(g,G_k)]$ is $[(g^{-1},G_k)]$.

The group thus constructed is called the "direct limit" or "directed limit" of the $G_n$ (though it should really be the direct limit of the system of groups-and-homomorphisms $(G_n, {f_{ij}})$), denoted by $$\lim_{\rightarrow} G_n.$$

More generally, the above arguments works for any system of groups-and-homorphisms $G_i$, $i\in I$, where $I$ is a directed partially ordered set ($I$ is a partially ordered set, and for every $a,b\in I$ there exists $c\in I$ such that $a\leq c$ and $b\leq c$), and for every $i\leq j$ we have a homomorphism $f_{ij}\colon G_i\to _Gj$, satisfying

• $f_{ii} = \mathrm{id}_{G_i}$ for each $i\in I$; and
• If $i\leq j\leq k$, then $f_{ik} = f_{jk}\circ f_{ij}$.

Such a system is called a "directed family of groups". We have:

Theorem. Let $(G_i, \{f_{ij}\})$ be a directed family of groups with directed index set $I$. Let $G=\lim_{\rightarrow} G_i$ be the directed limit of the family. Then:

1. For every $i\in I$ there is a group homomorphism $f_i\colon G_i\to G$ such that for all $i\leq j$, $f_i = f_j\circ f_{ij}$.
2. If $H$ is any group for which there exist homomorphisms $\phi_i\colon G_i\to H$ such that for all $i\leq j$, $\phi_i = \phi_j\circ f_{ij}$, then there exists a unique group homomorphism $\Phi\colon G\to H$ such that for all $i\in I$, $\phi_i = \Phi\circ f_i$.

Proof. Define $f_i\colon G_i\to G$ by $f_i(g)= [(g,G_i)]$. Since $[(g,G_i)] = [(f_{ij}(g),G_j)] = f_j(f_{ij}(g))$ for all $j\geq i$, condition 1 is satisfied.

Now suppose that $H$ and $\phi_i$ are given. Define $\Phi\colon G\to H$ by $\Phi([(g,G_n)]) = \phi_n(g)$.

This is well-defined: if $[(g,G_n)] = [(h,G_m)]$, then there exists $k\geq n,m$ such that $f_{nk}(g) = f_{mk}(h)$. Then $$\Phi[(h,G_m)] = \phi_m(h) = \phi_k(f_{mk}(h)) = \phi_k(f_{nk}(g)) = \phi_n(g) = \Phi[(g,G_n)].$$ It's not hard to verify that the map is a homomorphism, and is uniquely determined by this condition. QED

In the case at hand, you have the family indexed by the natural numbers, and the structure maps $f_{ij}$ are all embeddings. That means that you can think of the sequence of groups as mapping the groups you have so far into larger and larger and larger groups; the direct limit of the family is then the "limit" in the intuitive sense of that process: you simply take the union of all these nested groups, and make it into a group in the "obvious" way. However, because it is a directed limit, then you necessarily have the appropriate universal property.

So, suppose you have two families indexed by the same set, $(G_n,\{f_{ij}\})$ and $(H_n,\{g_{ij}\})$, and suppose moreover that, as in Luboš's answer, you have isomorphisms $\varphi_i\colon G_i\to H_i$ for each $i$. Then we have a situation like $$\begin{array}{ccccccccccc} G_1 & \stackrel{f_{12}}{\to} & G_2 & \stackrel{f_{23}}{\to} & G_3 & \stackrel{f_{34}}{\to} & \cdots & \stackrel{f_{n-1,n}}{\to} & G_n & \stackrel{f_{n,n+1}}{\to}&\cdots\\ \downarrow & &\downarrow & &\downarrow & &\cdots & & \downarrow & &\cdots\\ H_1 & \stackrel{g_{12}}{\to} & H_2 & \stackrel{g_{23}}{\to} & H_3 & \stackrel{h_{34}}{\to} & \cdots & \stackrel{g_{n-1,n}}{\to} & H_n & \stackrel{g_{n,n+1}}{\to}&\cdots \end{array}$$ Composing the maps $g_n\colon H_n\to \lim\limits_{\rightarrow}H_n$ with $\varphi_n\colon G_n\to H_n$. we get maps from each $G_i$ to $\lim\limits_{\rightarrow} H_n$, so the universal property of $\lim\limits_{\rightarrow}G_n$ gives a map $$\lim_{\rightarrow}G_n \longrightarrow \lim\limits_{\rightarrow}H_n.$$ If you compose the maps $f_n\colon G_n\to\lim\limits_{\rightarrow}G_n$ with the maps $\varphi_n^{-1}\colon H_n\to G_n$, then you get a map in the other direction by the universal property of $\lim\limits_{\rightarrow} H_n$. These two maps are inverses of each other (just check what happens to $[(g,G_n)]$ and what happens to $[(h,H_m)]$), so that the directed limits are isomorphic. The isomorphism is a consequence of the universal property of these constructions.

Morally, you want to think of the directed limit as a way of gluing together some partial information you might have among a large family of groups into a coherent whole. Here, the "partial information" consists of description of a certain nested family of groups (which will eventually become a nested family of subgroups of your "limiting group"), which you are gluing together in the obvious way. The construction becomes more interesting when the maps $f_{ij}$ are not necessarily embeddings.

-
How do you decide which statements are (merely?) propositions, and which are theorems? – The Chaz 2.0 May 26 '11 at 19:01
@The Chaz: Haphazardly; but "propositions" tend to be isolated statements, as opposed to bits that fit into a coherent theory. The fact that the multiplication is well defined is important, but it's a bit of a dead end in so far as it goes, so not really a landmark of the subject (a "theorem"). Could be a "lemma", I guess. – Arturo Magidin May 26 '11 at 19:10

The construction is called the direct limit.

-

The group is just the large $n$ limit of $C_{p^n}$. The group $C_{p^n}$ is clearly isomorphic to the multiplicative group $H_n$ of the numbers of the form $$z = \exp(2\pi i N / p^n)$$ where $N$ is an integer. Note that $z^{p^n}=1$. All these complex numbers lie on the unit circle and they're $p^n$-th roots of unity. The $n\to\infty$ limit of $C_{p^n}$ is therefore isomorphic to the limit of $H_n$, and it's defined by the description you wrote.

Note that the isomorphism of $C_{p^n}$ and $H_n$ defined above works in such a way that $\phi(x)$ for $x\in C_{p^n}$ is an element of $C_{p^{n+1}}$ which is, in the isomorphism above, mapped to the same element of $H_n$ or $H_{n+1}$ (the same complex number with unit norm) as $x$ itself.

-
It is not a limit; it is a direct limit, or colimit. – Qiaochu Yuan May 26 '11 at 17:47
Dear @Qiaochu, you would probably have to say much more to defend the idea that this "direct" adjective is more than just a meaningless complication in mathematical terminology. Is there also something that is called a limit of the sequence of groups? If not, I am afraid that the object described in the question is not a direct limit because it agrees neither in the detailed construction nor in the notation with the things you may find in definitions, e.g. on Wikipedia. – Luboš Motl May 26 '11 at 18:36
@Luboš: I don't understand your last sentence. This is the third example of a direct limit given in the Wikipedia article. I agree that "direct limit" is a bad term, but it's confusing to use the term "limit" in this context when there are closely related words relevant to this situation, some of which accurately describe it and some of which don't. – Qiaochu Yuan May 26 '11 at 18:40
@Lubos: The Prufer group is most certainly a direct (directed) limit (a categorical colimit). As for other kinds of "limits" that "sequence of groups" might have, there's the inverse limit, which in the case of the cyclic groups of prime power order would give you the $p$-adic integers, an animal altogether different from the Prufer group. Of course, it's not enough to consider the "sequence of groups", the connecting homomorphisms are an integral part of the constructions. – Arturo Magidin May 26 '11 at 18:49
Right, @Arturo, as long as you want to map particular elements of the groups in the sequence to the elements in the limiting group. While the question defines $\phi$ to allow such things, I think that what the question asks for doesn't require to map the individual elements. And the symbol of isomorphism, with the simple $\infty$ on the left hand side, doesn't really indicate that they're constructing any "special" limit. – Luboš Motl May 29 '11 at 16:18