Definition 2.2.10
Let \(a\) be an element of a group \(G\). The set \(\langle a \rangle = \{a^k \ : \ k \in \mathbf{Z}\}\) of powers of \(a\) is called the cyclic subgroup generated by \(a\). If there is an element \(a \in G\) such that \(\langle a \rangle = G\), we say that \(G\) is a cyclic group. We say that \(a\) is a generator of the cyclic group


Proposition 2.2.20
Let \(a\) be an element of a group \(G\).
  1. If \(a\) has infinte order, then \(\langle a \rangle\) is isomorphic to \(\mathbf{Z}\)
  2. If \(a\) has finite order, then \(\langle a \rangle\) is isomorphic to the group \(\mathbf{Z}_n\)


Proof
For \((a)\), we want to show this by finding an example of a isomorphism. So define the map \(\varphi \ : \ \mathbf{Z} \rightarrow \langle a \rangle\) by \(\varphi(k) = a^k\). To show that this map is an isomorphism, we need to show that it is a bijection and also that for any two elements \(a, b \in \mathbf{Z}\), \(\varphi(ab) = \varphi(a)\varphi(b)\) (Definition 2.1.13).

To show that it is a bijection, observe that this map is surjective or onto by the definition of \(\langle a \rangle\). (Recall that that \(\langle a \rangle\) is of infinite order). It is also injective or 1-1 because all the elements of \(\langle a \rangle\) (powers of \(a\)) are distinct. Furthermore, see that \(\varphi(k + l) = a^{k+l} = a^ka^k\). So \(\varphi\) is an isomorphism.

Similarly for \((b)\), we want to define an isomorphism. Note here that both \(\mathbf{Z}_n\) and \(\langle a \rangle\) have \(n\) elements so define the map \(\varphi \ : \ \mathbf{Z}_n \rightarrow \langle a \rangle\) by \(\varphi([k]) = a^k\) where \(0 \leq k \leq n-1\). In \(\mathbf{Z}_n\), the addition of \([k]\) and \([l]\) is \([r]\) where \(r\) is the remainder after dividing \(k+l\) by \(n\). While the multiplication in \(\langle a \rangle\) is given by \(a^ka^l = a^{k+l} = a^r\) where \(r\) is also the remainder after dividing \(k+l\) by \(n\) as we saw earlier. Therefore \(\varphi\) is an isomorphism.



Subgroups of Cyclic Groups

Every cyclic group is isomorphic to \(\mathbf{Z}\) or \(\mathbf{Z}_n\). Therefore, we’ll determine the subgroups of \(\mathbf{Z}\) and \(\mathbf{Z}_n\) to determine all the subgroups of cyclic groups.

Proposition 2.2.21
  1. Let \(H\) be a subgroup of \(\mathbf{Z}\). Then either \(H = \{0\}\) or there is a unique \(d \in \mathbf{N}\) such that \(H = \langle d \rangle = d\mathbf{Z}\)
  2. If \(d \in \mathbf{N}\), then \(d\mathbf{Z} \cong \mathbf{Z}\).
  3. If \(a, b \in \mathbf{N}\), then \(a\mathbf{Z} \subseteq b\mathbf{Z}\) if and only if \(b\) divides \(a\).


As a reminder, \(d\mathbf{Z} = \{ dn \ | \ n \in \mathbf{Z}\}\). It is the subgroup generated by \(d\).

Proof
For \((c)\):
\(\Rightarrow\): If \(a\mathbf{Z} \subseteq b\mathbf{Z}\), then \(a\) is a multiple of \(b\). So \(b\) divides \(a\).
\(\Leftarrow\): If \(b\) divides \(a\), then \(a \in b\mathbf{Z}\) so \(a\mathbf{Z} \subseteq b\mathbf{Z}\).

For \((a)\):
Suppose \(H\) is a subgroup of \(\mathbf{Z}\). Since \(H\) is a subgroup, then it must contain at least one element. Suppose \(H \neq \{0\}\), then \(H\) contains a nonzero integer. For any integer \(a \in H\), \(H\) must also contain \(-a\). Let \(d\) be the smallest element in \(H \cap \mathbf{N}\). We claim that \(H = d\mathbf{Z}\). We want to show that \(H \subseteq d\mathbf{Z}\) and \(d\mathbf{Z} \subseteq H\).

Let \(h \in H\). If \(h = d\), then since \(d \in H\), we have \(\langle d \rangle = d\mathbf{Z} \subseteq H\). Otherwise, we can write \(h = qd + r\) where \(0 \leq r < d\). Now, we know that \(h \in H\) and we know that \(qd \ \in H\) since it’s a multiple of \(d\), so \(r = h + (- qd) = h - qd \in H\). But \(d\) is the least positive element of \(H\) and \(r < d\). So we have \(r = 0\). Therefore, \(h = qd \in d\mathbf{Z}\).

So we know that there exists a \(d \in \mathbf{N}\) such that \(d\mathbf{Z} = H\). To see that it’s unique, suppose that \(d' \in H\) and so \(d'\mathbf{Z} = H\) but by (c), \(d\) and \(d'\) divide one another and since they’re both positive then \(d = d'\) so \(d\) is unique. \(\ \blacksquare\).

Corollary 2.2.22
Every subgroup of \(\mathbf{Z}\) other than \({0}\) is isomorphic to \(\mathbf{Z}\).



Lemma 2.2.23
Let \(n \geq 2\) and let \(d\) be a positive divisor of \(n\). The cyclic subgroup \(\langle [d] \rangle\) generated by \([d]\) in \(\mathbf{Z}_n\) has cardinality \(|\langle [d] \rangle| = n/d\)


Reminder: The subgroup generated by \(d\) is \(\{...,-2d,-d,0,d,2d,...\}\). The order of this subgroup is clearly infinite. The subgroup generated by \([d]\) is \(\{[0], [d], [2d],...,(k-1)[d]\}\) where \((k+1)[d] = 0\). This group is clearly finite.

Example: Suppose \(d=3\) and \(n=9\). \(Z_9 = \{0,1,2,3,4,5,6,7,8\}\) The cyclic subgroup in \(Z_9\) generated by \(\langle [3] \rangle\) is \(\{k[3] \mod 9 \ | \ k \in \mathbf{Z}\} = \{[0],[3],[6]\}\). The size of \(\langle [3] \rangle\) is \(n/d = 9/3 = 3\).

Proof
We are given that \(d\) is a positive divisor of \(n\). Since we’re working with addition, this means that for some multiple of \(d\), we will get back to \([0]\). So let \(s\) be the least positive integer such that \(s[d] = [0]\). In other words, \(sd \equiv 0 \mod n\). This implies that \(sd\) is a multiple of \(n\) so \(sd = kn\) for some integer \(k\). But since we need the smallest \(s\), then \(s = \frac{n}{d}\). \(\ \blacksquare\)

Proposition 2.2.24
Let \(H\) be a subgroup of \(\mathbf{Z}_n\).
  1. Either \(H = \{0\}\), or there is a \(d > 0\) such that \(H = \langle [d] \rangle\)
  2. If \(d\) is the smallest of positive integers \(s\) such that \(H\) = \(\langle [s] \rangle\), then \(d|H| = n\)


Example \((a)\): Let \(H\) be a subgroup of \(Z_9\). \(H\) must be generated by some \(d > 0\). For example \(Z_9\) is generated by \(H = \langle [1] \rangle\). \(\{[0],[3],[6]\}\) is generated by \(\langle [3] \rangle\) and so on.

Example \((b)\): Take \(s = 6\) such that \(H = \langle [6] \rangle = \{[0], [6], [3]\}\). But \(d = 3\) is the smallest generator for this subgroup because \(\langle [3] \rangle = \{[0], [3], [6]\}\).



Proof
For (a), suppose \(H \neq \{[0]\}\). Then \(H\) contains at least one equivalent such that it is not \([0]\). Let \([d]\) be the smallest positive equivalent class. We claim that \(H = \langle [d] \rangle\). So we need to show that \(\langle [d] \rangle \subseteq H\) and \(H \subseteq \langle [d] \rangle\).

\(\langle [d] \rangle \subseteq H\): Since \([d] \in H\), then by definition, \(\langle [d] \rangle \subseteq H\).

\(H \subseteq \langle [d] \rangle\). Suppose \([h] \in H\), then by the division algorithm we can find \(q\) and \(r\) such that

$$ \begin{align*} [h] = [qd] + [r] \end{align*} $$

We know that \([h] \in H\) and \([qd] \in H\). Since \(H\) is closed under addition, then

$$ \begin{align*} [r] = [h] - [qd] \end{align*} $$

is also in \(H\). But since \([d]\) was the smallest positive equivalent class, then \([r]=0\) and we must have \([h] = [qd]\). Therefore, \([h] \in \langle [d] \rangle\). From this we see that \(H = \langle [d] \rangle\) as we wanted to show.

For \((b)\), suppose that \(H = \langle [s] \rangle\) and \(d\) is the smallest of the positive integers \(s\) so \([d]\) also generates \(H\). By the division algorithm, write \(n = qd + r\) where \(0 \leq r < d\). Now

$$ \begin{align*} [n] &= [qd + r] \\ 0 &= [qd] + [r] \\ [r] &= -[qd] \\ &= -q[d] \\ \end{align*} $$

But this means that \([r] \in \langle [d] \rangle\)

Corollary 2.2.25
For a natural number \(n \geq 2\).
  1. Any subgroup of \(\mathbf{Z}_n\) is cyclic.
  2. Any subgroup of \(\mathbf{Z}_n\) has cardinality dividing \(n\).



Corollary 2.2.26
For a natural number \(n \geq 2\).
  1. For any positive divisor \(q\) of \(n\), there is a unique subgroup of \(\mathbf{Z}_n\) of cardinality \(q\) namely \(\langle [n/q] \rangle\).
  2. For any two subgroups \(H\) and \(H'\) of \(\mathbf{Z}_n\), we have \(H \subseteq H' \Leftrightarrow |H|\) divides \(|H'|\).


Example: Suppose \(n = 8\) and \(q = 2\). Then, there is a unique subgroup of cardinality \(q = 2\), namely \(H = \langle [n/q] \rangle = \langle [4] \rangle = \{[0],[4]\}\)

Proof
For \((a)\), since \(q\) is a positive divisor of \(n\), then by Lemma 2.2.23, the cyclic subgroup \(\langle [n/q] \rangle\) of \(\mathbf{Z}_n\) has cardinality \(n/(n/q) = q\). On the other hand, if \(H\) is a subgroup of cardinality \(q\), then by Proposition 2.2.24 \((b)\), …. TODO

Corollary 2.2.28
Let \(b \in \mathbf{Z}, b \neq 0\)
  1. The cyclic subgroup \(\langle [b] \rangle\) of \(\mathbf{Z}_n\) generated by \([b]\) is equal to the cyclic subgroup generated by \([d]\), where \([d] = gcd(b,n)\).
  2. The order of \([b]\) in \(\mathbf{Z}_n\) is \(\frac{n}{gcd(b,n)}\).
  3. In particular, \([b] = \mathbf{Z}_n\) if and only if \(b\) is relatively prime to \(n\).


Example: TODO

Proof

Proposition 2.2.30
Every subgroup of a cyclic group is cyclic.


Example: TODO

Proof

Proposition 2.2.31
Let \(a\) be an element of finite order \(n\) in a group. Then \(\langle a^k \rangle = \langle a \rangle \), if and only if \(k\) is relatively prime to \(n\). The number of generators of \(\langle a \rangle\) is \(\phi(n)\).


Example: TODO

Proof

Proposition 2.2.32
Let \(a\) be an element of finite order \(n\) in a group. For each positive integer \(q\) dividing \(n\), \(\langle a \rangle\) has a unique subgroup of order \(q\).


Example: TODO

Proof





References