Cyclic Groups

Now we return to the case of $G = \langle a\rangle$, a cyclic group. Clearly, $G = \{a^n\ \vert\ n \in \mathbb{Z}\}$. If $G$ is finite, then of course $o(a)$ must be finite, say $o(a) = n$. Then recalling our discussion of elements of finite order (see Property 10 from Chapter 2),


Thus we see $\vert G\vert = o(a)$ in this case.

Next, let $G_1 = \langle a\rangle$ and $G_2 = \langle b\rangle$ be two infinite cyclic groups. Consider the mapping given by $f(a^n) = b^n$ ($n=0,1,2,...$). This mapping is well-defined since $G_1$ is an infinite cyclic group generated by $a$ (all powers of $a$ are distinct since $G_1$ is infinite) and f is onto $G_2$. Also, so $f$ preserves the operation. Finally if $f(a^n) = f(a^m)$ then $b^n = b^m$. Since $G_2$ is infinite cyclic, generated by $b$, however, this implies $n = m$; hence $a^n = a^m$. Thus $f$ is an isomorphism mapping $G_1$ onto $G_2$ and so $G_1 \cong G_2$. We have shown that any two infinite cyclic groups are isomorphic. We note that since $\mathbb{Z}$, the additive group of integers, is infinite cyclic generated by $\pm 1$, we have proven any infinite cyclic group is isomorphic to $\mathbb{Z}$.

Next suppose that $G_1 = \langle a\rangle$ and $G_2 = \langle b\rangle$ are finite cyclic groups such that $\vert G_1\vert = \vert G_2\vert = n$. Then $G_1 = \{e, a, a^2,..., a^{n- 1}\}$, and $G_2 = \{e, b, b^2, ..., b^{n- 1}\}$. Define again by $f(a^k) = b^k$. To show that $f$ is well-defined, we note that if $a^t = a^m$, with $t \geq m$ say, then $a^{t-m} = e$ and therefore $o(a) = n\vert (t - m)$ (again see Property 10 Chapter 2). Therefore $b^{t- m} = e$, since $o(b) = n\vert (t - m)$, and $b^t = b^m$. Thus $f$ is well-defined. The rest of the justification needed to show $f$ is an isomorphism of $G_1$ ontp $G_2$ is left as an exercise (see exercise 1 for this section). Hence $G_1 \cong G_2$. This shows that any finite cyclic group of order $n$ is isomorphic to $\mu_n$ (see Example 5.1.4).

As our next consideration, we wish to determine the subgroups of a cyclic group. Suppose that $G = \langle a\rangle$ is a cyclic group and let $H \not= \{e\}$ be any subgroup of $G$. Let $t$ be the smallest positive integer such that $a^t \in H$; such a $t$ exists because $H$ must contain at least one positive power of $a$. For if $a^k \in H$ and $a^k \not= e$ then either $k > 0$ or $k < 0$. If $k > 0$, we have a positive power of $a$ in $H$, i.e., $a^k$. If $k < 0$, then $a^{- k} = (a^k)^{- 1}\in H$ and again we have a positive power of $a$ in $H$, i.e., $a^{- k}$. We claim that $H = \langle a^t\rangle$. For suppose that $x \in H$. Then $x = a^m$ for some integer $m$ since $H \leq G = \langle a\rangle$. By the Division Algorithm, $m = qt + r$, $0 \leq r < t$. Then $a^m = a^{qt}a^r$ and since $a^r= a^m a^{- qt}\in H$. This implies by the minimality of $t$, that $r = 0$ and $m = qt$. So $x = a^m = (a^t)^q$, i.e., $H = \langle a^t\rangle$. Therefore we have shown that any subgroup of a cyclic group is also cyclic.

Now suppose that $G = \langle a\rangle$ is an infinite cyclic group. We claim that $G$ has infinitely many subgroups. If $t$ is any positive integer, then $H = \langle a^t\rangle = \langle a^{-t}\rangle \leq G$. Let $t_1 \not= t_2$ be positive integers. If $\langle a^{t_1}\rangle=\langle a^{t_2}\rangle$ then we could write $a^{t_2}=a^{mt_1}$ for some $m \in\mathbb{Z}$ and and $a^{t_1}=a^{nt_2}$ for some $n \in\mathbb{Z}$. Hence $a^{t_2}=a^{mt_1}=a^{mnt_2}$. But $o(a) = \infty$, thus $m,n = \pm 1$, and $a^{t_2}=a^{\pm t_1}$. Again since $a$ has infinite order and since $t_1$ and $t_2$ are positive, we must have $t_2 = t_1$. The only way to avoid this contradiction is to conclude that the hypothesis $\langle a^{t_1}\rangle=\langle a^{t_2}\rangle$ is false. In other words, we have shown that if $t_1 \not= t_2$ then $\langle a^{t_1}\rangle\not=\langle a^{t_2}\rangle$. This proves the claim.

Suppose next that $G = \langle a\rangle$, and $\vert G\vert = n$. If $H \not= \{e\}$ is a subgroup, then we know, $H = \langle a^t\rangle$, where $t$ is the smallest positive integer such that $a^t \in H$. We claim that in this case $t\vert n$. For $a^n = e \in H$, so $n = qt$ by our earlier argument on the minimality of $t$ (in the paragraph above where we showed any subgroup of a cyclic group is also cyclic). Conversely if $t$ is a positive integer such that $t\vert n$, then $H =\langle a^t\rangle\leq G$. Moreover if $t_1$ and $t_2$ are positive integers such that $t_1 \not= t_2$ and $t_1\vert n$ and $t_2\vert n$ then $\langle a^{t_1}\rangle\not=\langle a^{t_2}\rangle$. For otherwise (i.e., assume instead that $\langle a^{t_1}\rangle=\langle a^{t_2}\rangle$), we must have $a^{t_1}=a^{qt_2}$ and $a^{t_2}=a^{kt_1}$. Thus $n\vert (t_1 - qt_2)$, but $t_2\vert n$ so $t_2\vert (t_1 - qt_2)$, so $t_2\vert t_1$. Similarly $t_1\vert t_2$ and so since $t_1,
t_2 > 0$ we must have $t_1 = t_2$, a contradiction.

Summarizing, we see that for a finite cyclic group $G = \langle a\rangle$, with $o(a) = n$, then any subgroup $H$ of $G$ is of the form $\langle a^t\rangle$, where $t > 0$ and $t\vert n$. Moreover, for each positive $t$ dividing $n$, there is such a subgroup. Also distinct divisors determine distinct subgroups. Finally since for a positive divisor $d$ of $n$ one has $(a^t)^d=e$ if and only if $n\vert td$ if and only if $(n/t)\vert d$, we conclude from Property 10 that $\vert H\vert=o(a^t)= n/t$, as $t$ runs through the positive divisors of $n$. Of course, if $t$ goes through all the positive divisors of $n$, so does $n/t$. Also note that the trivial subgroup $\{e\} = \langle a^0\rangle$. Thus in the case of cyclic groups of finite order, the converse of Lagrange's theorem is true, viz, for each $d\vert n$, $d > 0$, there exists a (cyclic) subgroup of order $d$. Moreover, in this case there is precisely one such subgroup (see exercise 2 for this section).

The converse of Lagrange's Theorem is not true in general, i.e., if $\vert G\vert = n$ and $d\vert n$, $d > 0$, there need not exist a subgroup of order $d$. The smallest example is the group $A_4$ of order $12$; it turns out that $A_4$ has no subgroup of order $6$. This will be left as an exercise (see exercise 4 for section 6.2) after more tools are developed to handle it efficiently. We shall meet further instances of this and will point it out for specific cases later. We point out now that it was no accident that the divisor was taken to be composite, i.e., not a prime, for prime divisors or for prime power divisors for that matter, there must exist subgroups of such orders. These matters will be attended to when we discuss the Sylow theorems.

Let us now summarize our results for cyclic groups.

Theorem 5.2.1  
Any two infinite cyclic groups are isomorphic (to $\mathbb{Z}$).

Any two finite cyclic groups of the same order are isomorphic (to some $\mu_n$, $n\geq 1$ - see Example 5.1.4).

Any subgroup of a cyclic group is cyclic. In particular if $H \leq G = \langle a\rangle$ and $H \not= \{e\}$, then $H = \langle a^t\rangle$, where $t$ is the smallest positive integer such that $a^t \in H$.

For an infinite cyclic group $G = \langle a\rangle$, and for each positive $t$, $H =\langle a^t\rangle\leq G$, and distinct positive $t$'s determine distinct subgroups.

If $G = \langle a\rangle$ is a finite cyclic group of order $n$, then the $t$ in part (c) divides $n$. Moreover to each positive divisor of $n$ there is one and only one cyclic subgroup of that order.

We shall see presently that (e) really characterizes finite cyclic groups, but first we establish an extremely useful theorem, which will be used many times. L

Theorem 5.2.2   Let $G$ be an arbitrary group and let a $G$ such that $o(a) = m$. Then $o(a^k)=m/gcd(m,k)$.

Proof: Let $t = o(a^k)$. Then $a^{kt} = (a^k)^t = e$ which implies that $m\vert kt$ (see Property 10 of elementary properties of groups, in chapter 2). Now write (see Corollary 1.2.7)

m=gcd(m,k)m',\ \ \ \

where $gcd(m' , k' ) = 1$. Hence,

gcd(m, k)m'\vert gcd(m, k)k' t,

or $m'\vert k' t$. because of Theorem 1.2.8 and the fact that $gcd(m' , k' ) = 1$, we have $m'\vert t$. But $m'=m/gcd(m,k)$, so $\frac{m}{gcd(m,k)} \vert t$. Now $(a^k)^{m/gcd(m,k)}=(a^m)^{k/gcd(m,k)}=e$. Thus $t\vert( m/(gcd(m, k))$. Therefore combining these two results, $o(a^k) = t =m/gcd(m, k)$. $\Box$

We apply the theorem immediately to the case of a finite cyclic group $G$ of order $n$. Let $G = \langle a\rangle = \{ e, a,..., a^{n- 1}\}$. By the theorem $o(a^k) = n/gcd(n, k)$. Thus $o(a^k) = n$ (and $a^k$ is consequently also a generator of $G$) if and only if $gcd (n, k) = 1$. In other words, if $G$ is a cyclic group of order $n$, then $G$ has $\phi(n)$ generators, where $\phi$ is the Euler $\phi$-function. Summarizing, we have the following result.

Corollary 5.2.3   Suppose $G = \langle a\rangle$ and $\vert G\vert = n$. Then $a^k$ is a generator of $G$ if and only if $k$ and $n$ are relatively prime. Thus there are $\phi(n)$ generators of $G$.

Using the same notation as before, for $G$ a cyclic group with $\vert G\vert = n$, let us see which elements, $a^k$, of $G$ are of order $d$. Again by Theorem 5.2.4, $o(a^k) = d$ if and only if $d = n/gcd(n, k)$. Thus $gcd(n, k) = n/d$. In which case, we can write $n=\frac{n}{d}d$, $k=\frac{n}{d}i$, where $gcd(d,i)=1$. Thus the elements of order $d$ are those of the form $a^{\frac{n}{d}i}$, where $gcd(i,d)=1$. and $0 < i \leq d - 1$ since $k \leq n - 1$. There are then, of course, $\phi (d)$ elements of order $d$. Since $\vert G\vert = n$, we have

\sum_{d\vert n} \phi(d) =n,
\end{displaymath} (5.1)

an interesting number theoretic relationship involving the $\phi$-function.

We shall make use of (5.1) directly in the following theorem. The author is indebted to W. Wardlaw for pointing this theorem out to him. (Also see [R].)

Theorem 5.2.4   If $\vert G\vert = n$ and if for each positive $d$ such that $d\vert n$, $G$ has at most one cyclic subgroup of order $d$, then $G$ is cyclic (and consequently, has exactly one cyclic subgroup of order $d$).

Proof: For each $d\vert n$, $d > 0$, let $\psi (d)$ = the number of elements of $G$ of order $d$. Then

\sum_{d\vert n} \psi(d) =n.

Now suppose that $\psi (d)\not= 0$ for a given $d\vert n$. Then there exists an $a \in G$ of order $d$ which generates a cyclic subgroup, $\langle a\rangle$, of order $d$, of $G$. We claim all elements of $G$ of order $d$ are in $\langle a\rangle$. Indeed, if $b \in G$ with $o(b) = d$ and $b \notin \langle a\rangle$, then $\langle b\rangle$ is a second cyclic subgroup of order $d$, distinct from $\langle a\rangle$. This contradicts the hypothesis, so the claim is proven. Thus, if $\psi (d)\not= 0$, then $\psi (d) = \phi(d)$. In general, we have $\psi (d) \leq \phi(d)$, for all positive $d\vert n$. But $n=\sum_{d\vert n}\psi (d) \leq\sum_{d\vert n} \phi(d)$, by our previous work. It follows, clearly, from this that $\psi (d) = \phi(d)$ for all $d\vert n$. In particular, $\psi (n) = \phi(n)\geq 1$. Hence, there exists at least one element of $G$ of order $n$; hence $G$ is cyclic. This completes the proof. $\Box$

Corollary 5.2.5   If in a group $G$ of order $n$, for each $d\vert n$, the equation $x^d = 1$ has at most $d$ solutions in $G$, then $G$ is cyclic.

Proof: The hypothesis clearly implies that $G$ can have at most one cyclic subgroup of order $d$ since all elements of such a subgroup satisfy the equation. So Theorem 5.2.4 applies to give our result. $\Box$

We have given a few applications along the way in this section and the preceding one of some of our results to number theory. We end this chapter with one further application. In particular, we claim that Theorem 4.3.6 is a generalization of Theorem 1.2.11, i.e., $gcd (a,b)\cdot lcm (a,b) = ab$. Let us see how this result follows from Theorem 4.3.6. We shall apply the theorem to the case where $G = \langle a\rangle$ is a cyclic group of finite order. This frequently, as we shall see in other instances, yields a number theoretic result as a special case of a group theoretical result. Let $s_1$, $s_2$ be arbitrary positive integers. Choose a positive integer $n$ such that $s_1\vert n$ and $s_2\vert n$. Then take $G = \langle a\rangle$ with $o(a) = n$. Let $H_1=\langle a^{s_1}\rangle$ and $H_2=\langle a^{s_2}\rangle$. Then from Theorem 5.2.2, $\vert H_1\vert=n/s_1$, and $\vert H_2\vert=n/s_2$. It is not difficult to show (see exercise 3 for this section) that

H_1\cap H_2= \langle a^{lcm(s_1,s_2)}\rangle,
\ \ \ \ {\rm and}
\ \ \ \
=\langle a^{gcd(s_1,s_2)}\rangle .

Thus Theorem 5.2.2 also implies that

\vert H_1\cap H_2\vert= n/lcm(s_1,s_2)
\ \ \ \ {\rm and}
\ \ \ \
\vert H_1H_2\vert=n/gcd(s_1,s_2).

Then Theorem 4.3.6 tells us that


which implies that $lcm (s_1, s_2) gcd(s_1, s_2) = s_1s_2$ as desired.


David Joyner 2007-08-06