On cyclic groups (or: a quick note from graduate school)

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Last week while reading Hungerford for my algebra class I got stuck on a theorem in an early section on groups where he proves that a subgroup generated by a finite subset $X$ (of a group $G$) consists of all finite products of the form $a_1^{n_1}\cdots a_k^{n_k}$ for all $x \in X$ and $n_i \in \mathbb{Z}$. After that was a similar statement for the cyclic subgroup $\langle a \rangle$ generated by a single element $a \in G$.

It was only after a while (read: an afternoon's worth of frustration) that I finally understood what was happening (my ‘proofreading’ skills — pun intended — still leave much to be desired). Up until then I had been exposed only to the converse of the statement: i.e., most books I’ve read have started by defining $\langle a \rangle$ as the set of all powers of an element $a \in G$ and then proving that it is a subgroup of $G$. Here, instead Hungerford starts by defining the subgroup $\langle a \rangle$ and then proceeding to show that it is in fact the set of all powers of $a$.

This episode reminded me of something my advanced calculus professor told us in the last term. Here I must provide a quick précis for the uninitiated: the set of real numbers $\mathbb{R}$ has a property that makes them of particular interest to mathematicians: it is complete. This means that every non-empty subset of $\mathbb{R}$ that is bounded above has a least upper bound or supremum. This property is not shared by the set of rational numbers $\mathbb{Q}$, because, for example, you are not going to find a rational number $q$ such that $q^2 = 2$. The remark was prompted by a question about why we are taking the existence of suprema axiomatically while having to prove the existence of infima as a consequence of the former. His answer (more or less) was that the choice of which existence is axiomatic is arbitrary in the sense that we could have chosen to axiomatize the existence of infima instead of suprema and then derive the existence of suprema from that. The reason why we don’t do this is perhaps partly because of convenience and partly because we have been so used to doing it that way. (This last bit might be an embellishment on my part.)

This insight was finally what helped me understand what Hungerford was doing. It felt good slowly piecing together these bits and pieces and feeling like I was finally starting to do ’real’ mathematics a couple of months into my first year of graduate school. And so I would like to use this post as an excuse to relish that moment and try discussing both ways of looking at cyclic groups (mainly as a memento for myself).

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Let’s start with the approach one would see in most introductory algebra texts. If $G$ is a group and $a \in G$ we define the set $\langle a \rangle$ as the set of all powers (or multiples) of $a$, i.e., $$\langle a \rangle = \{a^n : n \in \mathbb{Z}\}.$$ Notice that we do not claim anything about the structure of $\langle a \rangle$. Instead we simply define it to be a set. We come know to the following theorem:

Theorem 1. Let $G$ be a group and $a \in G$. Then $\langle a \rangle$ is a subgroup of $G$.

Proof. Since $a \in \langle a \rangle$ it follows that $\langle a \rangle \neq \emptyset$. Let $\xi$, $\eta$ be elements of $\langle a \rangle$. Then there exist integers $m,n$ such that $\xi = a^m$ and $\eta = a^n$ so that $$\xi\eta^{-1} = a^ma^{-n} = a^{m-n} \in \langle a \rangle .$$ Hence (by the one-step subgroup test) $\langle a \rangle$ is a subgroup of $G$. QED.

This is essentially the result we wanted. We can now proceed to the converse approach.

Definition 2. (Hungerford) Let $G$ be a group and $X$ a subset of $G$. Let $(H_\lambda)_{\lambda \in \Lambda}$ be the family of all subgroups of $G$ that contain $X$. Then the intersection $H = \bigcap_{\lambda \in \Lambda} H_\lambda$ is called the subgroup of $G$ generated by $X$ and is denoted by $\langle X \rangle$.

That the intersection $\bigcap H_\lambda$ is a subgroup of $G$ follows from Corollary 2.6 in Hungerford (p. 32). If $X$ is finite then we write $\langle X \rangle$ as $\langle x_1, \ldots, x_n \rangle$. If $X = \{a\}$ then we write $\langle a \rangle$ and call the latter the cyclic subgroup of $G$ generated by $a$.

Theorem 3. If $G$ is a group and $X$ a non-empty subset of $G$ then $\langle X \rangle$ consists of all finite products of the form $a_1^{n_1}\cdots a_k^{n_k}$ for all $x \in X$ and $n_i \in \mathbb{Z}$. In particular, if $a \in G$ then $\langle a \rangle$ consists of all powers of $a$.

Proof.Let $$H := \{a_1^{n_1}\cdots a_k^{n_k} : a_i \in X, n_i \in \mathbb{Z}, k \in \mathbb{N}\}.$$ The goal is to show that $$H \leq \langle X \rangle \leq H$$ so that $\langle X \rangle = H$. Let $\xi \in X$. Then $\xi \in H$ (setting $a_1 = \xi$, $n_1 = 1$, and $k = 1$). Hence $X \subseteq H$. Since $G$ is a group it follows that the binary operation in $G$ is associative. Moreover $\xi^0 = e \in H$. Finally if $a_1^{n_1}\cdots a_k^{n_k} \in H$ then so is $a_1^{-n_1}\cdots a_k^{-n_k}$ (i.e., every element of $H$ has an inverse). Therefore $\langle X \rangle \leq H$.

Conversely, let $H_\lambda$ be a subgroup of $G$ containing $X$. Then we have $a_1, \ldots, a_k \in X \subseteq H_\lambda$. Since $H_\lambda$ is a group we have $$a_1^{n_1}, \ldots, a_k^{n_k} \in H_\lambda$$ and furthermore $$a_1^{n_1}\cdots a_k^{n_k} \in H_\lambda$$ so that $H \leq H_\lambda$ for all $H_\lambda$. We then have $$H \leq \bigcap_{\lambda \in \Lambda} H_\lambda = \langle X \rangle$$ and so $H \leq \langle X \rangle \leq H$. Therefore $\langle X \rangle = H$. QED.

The definition at the start of this chapter, when viewed instead as a result, follows immediately from Theorem 3.

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We conclude by reviewing some properties of cyclic groups and subgroups that make them useful.

Theorem 4. Every subgroup of the additive group $\mathbb{Z}$ is cyclic.

Proof. Let $H$ be a subgroup of $\mathbb{Z}$. If $H = \{0\}$, then $H = \langle 0\rangle$ and we are done. Otherwise, let $m$ be the smallest positive integer in $H$. We want to show that $H = \langle m \rangle$.

If $m \in H$ then every multiple of $m$ is in $H$ since $H$ is closed under addition, and so $\langle m \rangle \leq H$. Now let $h \in H$. Then there exist $q,r \in \mathbb{Z}$ such that $h = mq + r$ and $0 \leq r \lt m$. Since $h \in H$ and $mq \in H$, $r = h - mq \in H$. But $r \lt m$ and $m$ is the smallest positive integer in $H$, so $r = 0$. Therefore $h = mq$ and $h \in \langle m \rangle$. Thus $H \leq \langle m \rangle$ and so $H = \langle m \rangle$.

That $\mathbb{Z}$ (by Theorem 4) and $\mathbb{Z}_n$ (by the definition of modular arithmetic) are cyclic groups leads us to the following result:

Theorem 5. Every infinite cyclic group is isomorphic to $\mathbb{Z}$ and every finite cyclic group of order $n$ is isomorphic to $\mathbb{Z}_n$.

Proof. Let us first consider the case where the cyclic group is infinite. Let $G = \langle a \rangle$ where $a$ is a generator of $G$. Define $\varphi : \mathbb{Z} \to G$ by $\varphi(n) = a^n$. We want to show that $\varphi$ is an isomorphism.

First we show that $\varphi$ is a homomorphism. Let $m,n \in \mathbb{Z}$. Then $$\varphi(m + n) = a^{m + n} = a^m a^n = \varphi(m) \varphi(n).$$ Therefore $\varphi$ is a homomorphism. Now we show that $\varphi$ is injective. Let $m,n \in \mathbb{Z}$ such that $\varphi(m) = \varphi(n)$. Then $a^m = a^n$ and $m = n$, so that $\varphi$ is injective. Finally we show that $\varphi$ is surjective. Let $g \in G$. Since $G = \langle a \rangle$, there exists $n \in \mathbb{Z}$ such that $g = a^n.$ Therefore $\varphi(n) = a^n = g$ and $\varphi$ is surjective. Since $\varphi$ is a bijective homomorphism, $\varphi$ is an isomorphism.

Now let us show that a finite group of order $n$ is isomorphic to $\mathbb{Z}_n$. (This proof is due to Hungerford.) For all integers $r$ and $s$ we have $$a^r = a^s \iff a^{r - s} = e \iff r-s \in \ker \varphi = \langle m \rangle$$ so that $$a^r = a^s \iff n \mid (r - s) \iff r \equiv s \pmod{n}.$$ If we denote by $[k]$ the equivalence class of $k$ modulo $n$ for all $k \in \mathbb{Z}$, then the map $\psi : \mathbb{Z}_n \to G$ defined by $\psi([k]) = a^k$ is a well-defined epimorphism. Moreover, $$\psi([k]) = e \iff a^k = e \iff n \mid k \iff [k] = [0]$$ and thus $\psi$ is injective. Therefore $\psi$ is an isomorphism and $G \cong \mathbb{Z}_n$. QED.

Theorem 6. (Fundamental theorem of cyclic groups) Every subgroup of a cyclic group is cyclic. Every homomorphic image of a cyclic group is cyclic.

Proof. The first statement can be demonstrated by reworking the proof of Theorem 4 using multiplicative notation. We will prove the second statement. Let $\varphi: G \to H$ be a group homomorphism and let $y \in \operatorname{im} \varphi$. Then there exists some element $x \in G$ such that $\varphi(a) = y$. Suppose that $G$ is cyclic and $a \in G$ is a generator. Then $x = a^n$ for some $n \in \mathbb{Z}$. Then $$\varphi(x) = \varphi(a^n) = \varphi(a)^n = y.$$ Since $y$ was arbitrary, we have $\operatorname{im} \varphi = \langle \varphi(a) \rangle$ and therefore, $\varphi(G)$ is cyclic.