1.4. Cyclic Groups.
Cyclics groups are a special type of group that are easy to recognize as group structures. In a cyclic group, one can always pinpoint a single element which can "generate" every other element of the group. For example, the subgroup of rotations \(\{e, r, r^2, \dots, r^{n-1}\}\) in dihedral groups is cyclic. In such a subgroup, every element is simply a fininte power of \(r\). We can then think of this subgroup as being generated by a single element, namely \(r\).
It will turn out later that, in every group, there will always be a subset of its elements such that the subset generates the whole group. Here, we're starting small, by just considering groups whose elements can be generated by one element.
A group \(G\) is cyclic if there exists an element \(g \in G\) such that
A very trivial example of a cyclic group is the integers under addition. This is because every element in \((\mathbb{Z}, +)\) can be generated by the number 1 (e.g., 3 = 1 + 1 + 1, -2 \(= -1 -1\)). With the definition of cyclic groups at hand, we can introduce theorems about cyclic groups.
Let \(G\) be a cyclic group, and suppose \(G = \{g^n | n \in \mathbb{N}\}\) for some \(g \in G\). Then \(|G| = |g|\).
Suppose \(|g| = k\) where \(k\) is some positive integer. Then since \(G = \{g^n | n \in \mathbb{N}\}\),
Therefore \(|G| = k = |g|\). Now if \(|g| = \infty\), then by the same exact reasoning \(|G| = \infty\). A consequence of this theorem is that if \(|g| = \infty\), then we know that \(g^a \ne g^b\) for any \(a, b \in \mathbb{N}\) such that \(a \ne b\). This would imply that \(g^{a - b} = 1\), which otherwise imply the group \(G\) to have finite order; a contradiction if \(|g| = \infty\) since this implies \(|G| = \infty\).
Any two cyclic groups of the same order are isomorphic.
Let \(G\) and \(G'\) be two cyclic groups and suppose \(|G| = |G'|\). Furthermore suppose that \(G = \left< g\right>\) and \(G' = \left< g' \right>\). Construct the homomorphism \(\phi: G \to G'\) where
for any \(n \in \mathbb{N}\). Observe that this is surjective, as the groups are of equal order so for any \((g')^n\) there we can identify the preimage to be \(g^n\). This is also injective, since if \(g_1' = g_2'\) are both elements in \(G'\), then \(g_1' = g_2' = (g')^n\) for some \(n\) which corresponds to one and only one element in \(G\); namely, \(g^n\). As the homomorphism we constructed is surjective and injective, we have that \(\phi\) is an isomorphism so that the groups are isomorphic.
Note we could have also utilized Theorem 1.3 here, by constructing the homomorphism \(\psi: G' \to G\) where \(\psi((g')^n) = g^n\).
We'll now move onto a more useful theorem concerning cyclic groups.
Let \(G\) be a cyclic group. Then every subgroup of \(G\) is cyclic.
Let \(H\) be a subgroup of \(G\). Let \(m\) be the smallest integer such that \(g^m \in H\). Suppose towards a contradiction that \(H\) is not cyclic. That is, there exists an element \(h \in H\) such that \(h \ne g^{mn}\) for any \(m \in \mathbb{N}\).
Since \(h \in G\), and \(G\) is a cyclic group, it must be some integer power of \(g\). Since \(h \ne g^m\) for any \(m \in \mathbb{N}\), we know that there must exist \(q, r \in \mathbb{Z}\) such that \(h = g^{qm + r}\) where \(0 < r < m\).
Now since \(H\) is a group, \(h^{-1} = g^{-(qm + r)} \in H\). Furthermore, \(g^{(q+1)m} \in H\) since \(H\) is closed under products. By the same reasoning,
is in \(H\). However, this contradicts our assumption that \(m\) is the smallest positive integer such that \(g^m \in H\). Thus by contradiction \(H\) must be cyclic. Remark. This proof utilizes the important idea that, if you know \(G\) is a group, and \(g_1, g_2\) are any two elements of \(G\), then the product \(g_1g_2 \in G\). Furthermore, if \(g_1, g_2, \dots, g_n \in G\), then \(g_1^{n_1}g_2^{n_2}\cdots g_k^{n_k} \in G\) for literally any powers \(n_1, n_2 \dots n_k \in \mathbb{Z}\).
We used this idea by (1) observing that \(g^m \in H\), so that \(g^{(q+1)m} \in H\) for any \(q \in \mathbb{Z}\) and (2) reasoning that \(g^{(q+1)m}g^{-(qm + r)} \in H\).
In addition, we'll offer a way to think about subgroups of a cyclic group \(G\):
In the above representations, the color-highlighted powers of \(g\) form subgroups of \(G\). Thus \(\{{\color{blue}e}, {\color{blue}g^2}, {\color{blue}g^4}, {\color{blue}g^6}, \dots\}\), \(\{{\color{red}e}, {\color{red}g^3}, {\color{red}g^6}, {\color{red}g^9}, \dots \}\) and \(\{{\color{magenta}e}, {\color{magenta}g^4}, {\color{magenta}g^8}, \dots\}\) are all subgroups of \(G\). Of course, this process will eventually terminate if \(G\) is finite, but we represent the most general case above with the \(\dots\) terms. In addition, if \(G\) is finite one of the subgroups may turn out to be just the entire group \(G\).
To find the exact subgroups of a cyclic group we can use the following theorem.
Let \(G\) be a finite cyclic group of order \(n\). For every positive integer \(d\) such that \(d|n\), there is exactly one subgroup of order \(d\) of \(G\). These are all the subgroups of \(G\).
Let \(H\) be a subgroup of \(G\) and suppose \(|H| = m \le n\). We'll first show that \(n \mid m\) and then show that for every \(d \mid n\), there exists a subgroup of order \(d\).
By the previous theorem, we know that \(H\) must be cyclic. Therefore \(H = \{h^n \mid n \in \mathbb{N}\}\) for some \(h \in G\). However, by Theorem 1.4, we also know that \(|H| = |h|\), so that \(h^m = e\). Therefore,
However, if \(G = \{e, g, g^2, \dots, g^{n-1}\}\) for some \(g \in G\), then \(h = g^k\) for some positive integer \(k < n\). Therefore
Since \(g^k\) generates \(H\), we can apply Theorem 1.4 to conclude that \(|g^k| = m\). In order for this to be true, there must exist some positive integer \(j\) such that \(g^{jk} = e\).
On one hand, we already know \(m\) is the smallest such integer, as we specified \(h^m = (g^k)^m = e\). On the other, we also know that \(|g| = n\); that is, \(n\) is the smallest integer such that \(g^n = e\). Therefore, we have that
which proves that \(m \mid n\).
To show that for each \(d \mid n\) there exists a subgroup of order \(d\), simply observe that if \(G = \{e, g, g^2, \dots , g^n\}\) then the set \(\{e, g^{n/d}, g^{2n/d}, \dots, g^{(d-1)n/d}\}\) is a subgroup of order \(d\). This completes the proof.