Subgroups of cyclic groups
The following two propositions determine every subgroup of every cyclic group.
Subgroup of \( \mathbb{Z} \)
Let \( G \) be a subgroup of \( \mathbb{Z} \). Then \( G = d\mathbb{Z} \) for some \( d \geq 0 \).
Subgroup of \( \mathbb{Z}/n\mathbb{Z} \)
Let \( G \) be a subgroup of \( \mathbb{Z}/n\mathbb{Z} \) for some \( n \geq 0 \). Then \( G = \mathbb{Z}/d\mathbb{Z} \) for some \( d \geq 0 \) with \( d \mid n \).
Can you recall the proofs?
Poof idea
Search for the smallest positive element in \( G \). Start by choosing any element \( b \in G \). Search for a smaller element by considering \( b - a \) for every \( a \in G \). Eventually, you will find an element \( c \in G \) that is both the smallest positive element in \( G \) and divides every other element in \( G \).
Proof. Let \( G \subseteq \mathbb{Z} \) be a subgroup. If \( G = \{0\} \), then \( G = 0\mathbb{Z} \). Otherwise, let \( b \in G \) be a nonzero element of \( G \). With \( b \in G \), we alse have \( -b \in G \), and so we can say that \( G \) contains positive integers. With positive integers in \( G \), we can ask for the smallest positive integer in \( G \), thanks to the well-ordering principle of \( \mathbb{Z} \). Let \( d \) be the smallest positive integer in \( G \). We will show that \( G = d\mathbb{Z} \). First, we have \( d\mathbb{Z} \subseteq G \), as \( d \) is in \( G \) and \( d\mathbb{Z} \) is the smallest subgroup of \( \mathbb{Z} \) containing \( d \). Second, we have \( G \subseteq d\mathbb{Z} \). To see this, let \( a \in G \). By the division algorithm, we can write \( a = qd + r \) for some integers \( q \) and \( r \) with \( 0 \leq r < d \). Since \( a \) and \( qd \) are in \( G \), we have \( r = a - qd \in G \). Since \( r \) is a positive integer less than \( d \), we must have \( r = 0 \), as \( d \) is the smallest positive integer in \( G \). Thus, \( a = qd \in d\mathbb{Z} \). Therefore, \( G \subseteq d\mathbb{Z} \), and we have \( G = d\mathbb{Z} \).
Proof idea
The same idea above can be repeated here. Aluffi, on the other hand, takes a more interesting approach that uses earlier results. He considers passing a subgroup back through the quotient map from \( \mathbb{Z} \) to \( \mathbb{Z}/n\mathbb{Z} \). The result must be a subgroup of \( \mathbb{Z} \), and so it must be of the form \( d\mathbb{Z} \), by the above proposition. Then, he shows that \( d \mid n \).
Proof. Let \( G \subseteq \mathbb{Z}/n\mathbb{Z} \) be a subgroup. Let \( \pi_n : \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z} \) be the quotient map. Then \( G' = \pi_n^{-1}(G) \) is a subgroup of \( \mathbb{Z} \), and so it is of the form \( d\mathbb{Z} \) for some \( d \geq 0 \). In other words, \( G' \) is generated by \( d \), denoted as \( G' = \langle d \rangle \). Going forward through the quotient map, we have \[G = \pi_n(G') = \pi_n(\langle d \rangle) = \langle [d]_n \rangle.\] Thus, \( G \) is generated by an element \( [d]_n \). In addition to this result, notice that \( n \in G' \), as \( \pi_n(n) = [0]_n \in G \), which means that the generator \( d \) of \( G' \) divides \( n \).