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Section4.2Subgroups of a Cyclic Group

We can ask some interesting questions about cyclic subgroups of a group and subgroups of a cyclic group. If \(G\) is a group, which subgroups of \(G\) are cyclic? If \(G\) is a cyclic group, what type of subgroups does \(G\) possess?

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Theorem4.2.1

Every subgroup of a cyclic group is cyclic.

##### Proof

The main tools used in this proof are the division algorithm and the Principle of Well-Ordering. Let \(G\) be a cyclic group generated by \(a\) and suppose that \(H\) is a subgroup of \(G\text{.}\) If \(H = \{ e \}\text{,}\) then trivially \(H\) is cyclic. Suppose that \(H\) contains some other element \(g\) distinct from the identity. Then \(g\) can be written as \(a^n\) for some integer \(n\text{.}\) Since \(H\) is a subgroup, \(g^{-1} = a^{n}\) must also be in \(H\text{.}\) Since either \(n\) or \(-n\) is postive, we can assume that \(H\) contains positve powers of \(a\) and \(n \gt 0\text{.}\) Let \(m\) be the smallest natural number such that \(a^m \in H\text{.}\) Such an \(m\) exists by the Principle of Well-Ordering.

We claim that \(h = a^m\) is a generator for \(H\text{.}\) We must show that every \(h' \in H\) can be written as a power of \(h\text{.}\) Since \(h' \in H\) and \(H\) is a subgroup of \(G\text{,}\) \(h' = a^k\) for some integer \(k\text{.}\) Using the division algorithm, we can find numbers \(q\) and \(r\) such that \(k = mq +r\) where \(0 \leq r \lt m\text{;}\) hence,

\begin{equation*}
a^k = a^{mq +r} = (a^m)^q a^r = h^q a^r.
\end{equation*}
So \(a^r = a^k h^{-q}\text{.}\) Since \(a^k\) and \(h^{-q}\) are in \(H\text{,}\) \(a^r\) must also be in \(H\text{.}\) However, \(m\) was the smallest positive number such that \(a^m\) was in \(H\text{;}\) consequently, \(r=0\) and so \(k=mq\text{.}\) Therefore,

\begin{equation*}
h' = a^k = a^{mq} = h^q
\end{equation*}
and \(H\) is generated by \(h\text{.}\)

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Corollary4.2.2

The subgroups of \({\mathbb Z}\) are exactly \(n{\mathbb Z}\) for \(n = 0, 1, 2,\ldots\text{.}\)

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Proposition4.2.3

Let \(G\) be a cyclic group of order \(n\) and suppose that \(a\) is a generator for \(G\text{.}\) Then \(a^k=e\) if and only if \(n\) divides \(k\text{.}\)

First suppose that \(a^k=e\text{.}\) By the division algorithm, \(k = nq + r\) where \(0 \leq r \lt n\text{;}\) hence,

\begin{equation*}
e = a^k = a^{nq + r} = a^{nq} a^r = e a^r = a^r.
\end{equation*}
Since the smallest positive integer \(m\) such that \(a^m = e\) is \(n\text{,}\) \(r= 0\text{.}\)

Conversely, if \(n\) divides \(k\text{,}\) then \(k=ns\) for some integer \(s\text{.}\) Consequently,

\begin{equation*}
a^k = a^{ns} = (a^n)^s = e^s = e.
\end{equation*}
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Theorem4.2.4

Let \(G\) be a cyclic group of order \(n\) and suppose that \(a \in G\) is a generator of the group. If \(b = a^k\text{,}\) then the order of \(b\) is \(n/d\text{,}\) where \(d = \gcd(k,n)\text{.}\)

##### Proof

We wish to find the smallest integer \(m\) such that \(e = b^m = a^{km}\text{.}\) By Proposition Proposition 4.2.3, this is the smallest integer \(m\) such that \(n\) divides \(km\) or, equivalently, \(n/d\) divides \(m(k/d)\text{.}\) Since \(d\) is the greatest common divisor of \(n\) and \(k\text{,}\) \(n/d\) and \(k/d\) are relatively prime. Hence, for \(n/d\) to divide \(m(k/d)\) it must divide \(m\text{.}\) The smallest such \(m\) is \(n/d\text{.}\)

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Corollary4.2.5

The generators of \({\mathbb Z}_n\) are the integers \(r\) such that \(1 \leq r \lt n\) and \(\gcd(r,n) = 1\text{.}\)

Let us examine the group \({\mathbb Z}_{16}\text{.}\) The numbers 1, 3, 5, 7, 9, 11, 13, and 15 are the elements of \({\mathbb Z}_{16}\) that are relatively prime to 16. Each of these elements generates \({\mathbb Z}_{16}\text{.}\) For example,

\begin{align*}
1 \cdot 9 & = 9 & 2 \cdot 9 & = 2 & 3 \cdot 9 & = 11\\
4 \cdot 9 & = 4 & 5 \cdot 9 & = 13 & 6 \cdot 9 & = 6 \\
7 \cdot 9 & = 15 & 8 \cdot 9 & = 8 & 9 \cdot 9 & = 1 \\
10 \cdot 9 & = 10 & 11 \cdot 9 & = 3 & 12 \cdot 9 & = 12\\
13 \cdot 9 & = 5 & 14 \cdot 9 & = 14 & 15 \cdot 9 & = 7.
\end{align*}