[2.2] Subgroup and Cyclic Groups
- If \(a\) has infinte order, then \(\langle a \rangle\) is isomorphic to \(\mathbf{Z}\)
- 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.
- Let \(H\) be a subgroup of \(\mathbf{Z}\). Then either \(H = \{0\}\) or there is a unique \(d \in \mathbf{N}\) such that \(H = \langle d \rangle = d\mathbf{Z}\)
- If \(d \in \mathbf{N}\), then \(d\mathbf{Z} \cong \mathbf{Z}\).
- 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. \(\ \blacksquare\).
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 \(s = \frac{n}{d}\). \(\ \blacksquare\)
- Either \(H = \{0\}\), or there is a \(d > 0\) such that \(H = \langle [d] \rangle\)
- 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
We know that \([h] \in H\) and \([qd] \in H\). Since \(H\) is closed under addition, then
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
But this means that \([r] \in \langle [d] \rangle\)
- Any subgroup of \(\mathbf{Z}_n\) is cyclic.
- Any subgroup of \(\mathbf{Z}_n\) has cardinality dividing \(n\).
- For any positive divisor \(q\) of \(n\), there is a unique subgroup of \(\mathbf{Z}_n\) of cardinality \(q\) namely \(\langle [n/q] \rangle\).
- 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
- The cyclic subgroup \(\langle [b] \rangle\) of \(\mathbf{Z}_n\) generated by \([b]\) is equal to the cyclic subgroup generated by \([d]\), where \([d] = gcd(b,n)\).
- The order of \([b]\) in \(\mathbf{Z}_n\) is \(\frac{n}{gcd(b,n)}\).
- In particular, \([b] = \mathbf{Z}_n\) if and only if \(b\) is relatively prime to \(n\).
Example: TODO
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Example: TODO
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Example: TODO
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Example: TODO
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