A cyclic

group **G** =
‹

*a* › generated by the element

*a* of

**G** is the set of all integral powers of

*a*, which are

*a*^{k} in multiplicative notation and k

*a* in additive notation.
If

**G** is a

group (not necessarily cyclic) and

*a* is an element of

**G**, then
‹

*a* › is the cyclic

subgroup of

**G** generated by

*a*, denoted ‹

*a* › ≤

**G**.

The cyclic subgroup of **G** generated by the element *a* is also the intersection of all subgroups **H** of **G** that contain *a*. Thus, the cyclic subgroup ‹ *a* › generated by the element *a* of **G** is the smallest subgroup of **G** that contains *a*.

**Theorem**: Every subgroup of a cyclic group is cyclic.

__Proof__:

Let **G** = ‹ *a* › be a cyclic group and let **H** be a subgroup of **G**.

Then, **H** is a subset of **G** and so, for any element *h* of **H**, *h* = *a*^{k} for some integer k.

If there does not exist an integer m such that *a*^{m} ≠ 1 (where 1 denotes, of course, the identity element of **G**) and *a*^{m} is an element of **H**, then **H** = {1}, and **H** is trivially a cyclic subgroup of **G**.

If there does exist an integer m such that *a*^{m} ≠ 1 and *a*^{m} is an element of **H**, then either m < 0 or m > 0.

If m < 0, then, since **H** is a group it contains the inverse of *a*^{m}, (*a*^{m})^{-1} = *a*^{-m}.

*a*^{m}*a*^{-m} = 1, and if *a*^{-m} = 1, then 1 = *a*^{m}*a*^{-m} = 1*a*^{m} = *a*^{m}, which is a contradiction, and so *a*^{-m} ≠ 1, and -m > 0, since m < 0.

Therefore, there exists a positive integer m such that *a*^{m} ≠ 1 and *a*^{m} is an element of **H**.

Let **S** = { n | n is a positive integer and *a*^{n} ≠ 1 and *a*^{n} is an element of **H**}.

Since **S** is non-empty, by the well-ordering of the positive integers (natural numbers), **S** has a least element n_{0}.

‹ *a*^{n0} › is a subgroup of **H** and therefore if *h* is an element of **H**, then *h* = *a*^{k}, where k = qn_{0} + r, for some integer q and an integer r such that 0 ≤ r < n_{0}. (See division algorithm for proof of this)

*a*^{qn0} = (*a*^{n0})^{q} is an element of ‹ *a*^{n0} ›, which is a group, and so its inverse (*a*^{qn0})^{-1} = *a*^{-qn0} is an element of

‹ *a*^{n0} › and also of **H** (since ‹ *a*^{n0} › is a subgroup of **H**).

**H** is a group, and so is closed its group operation, which is denoted here by juxtapostion.

Therefore,
*a*^{k}*a*^{-qn0} = *a*^{qn0 + r}*a*^{-qn0} = *a*^{r} is an element of **H**.

n_{0} is the least positive integer such that *a*^{n0} ≠ 1 and r < n_{0}, and so *a*^{r} = 1.

Therefore, *a*^{k} = *a*^{qn0 + r} = *a*^{qn0}*a*^{r} = *a*^{qn0}1 = *a*^{qn0} and *a*^{k} is an element of ‹ *a*^{n0} ›

Therefore, **H** is a subset of ‹
*a*^{n0} ›, which is itself a subset of **H**, and so **H** = ‹ *a*^{n0} › and **H** is cyclic. (QED)

The following are a few important results concerning finite cyclic groups.

**Theorem**: If **G** = ‹ *a* › is a cyclic group of order n and 1 ≤ k < n, then

**A**. The order of *a*^{k}, denoted **o**(*a*^{k}), which is the least positive integer m such that (*a*^{k})^{m} = 1,

is n/(n,k), where (n,k) is the greatest common divisor of n and k.

In particular, *a*^{k} generates **G** if and only if (n,k) = 1, i.e., n and k are relatively prime.

__Proof__:

Let d = (n,k).
Since d is the greatest common divisor of n and k, there exist integers x and y such that

d = kx + ny. (See greatest common divisor for proof of this.)

Then, *a*^{d} = *a*^{kx + ny} = *a*^{kx}*a*^{ny}

Since **G** = ‹ *a* › is a cyclic group of order n, *a*^{n} = 1, and *a*^{ny} = 1^{y} = 1.

Therefore, *a*^{d} = *a*^{kx + ny} = *a*^{kx}*a*^{ny} = *a*^{kx}1 = *a*^{kx} = (*a*^{k})^{x} and *a*^{d} ε ‹ *a*^{k} › , which implies that ‹ *a*^{d} › is a subset of ‹ *a*^{k} ›

d divides k, so there exists an integer m such that m = k/d and *a*^{k} = (*a*^{d})^{k/d} = (*a*^{d})^{m} and *a*^{k} ε ‹ *a*^{d} ›,

which implies ‹ *a*^{k} › is a subset of ‹ *a*^{d} ›

Since ‹ *a*^{k} › is a subset of ‹ *a*^{d} › and ‹ *a*^{d} › is a subset of ‹ *a*^{k} ›, ‹ *a*^{d} › = ‹ *a*^{k} ›

Therefore, **o**(*a*^{k}) = **o**(*a*^{d}), but d divides n and so the least integer m such that dm = n is n/d.

Therefore, **o**(*a*^{k}) = n/d = n/(n,k) (QED).

The second part of the claim follows easily. If **G** = ‹ *a* › is a cyclic group of order n, then **o**(*a*^{k}) = n if and only if (n,k) = 1, which is just to state that ‹ *a*^{k} › = **G**.

**B**. If d divides n, then **o**(*a*^{k}) = d if and only if k = r(n/d) and (r,d) = 1.

__Proof__:

From, part **A** we have that **o**(*a*^{k}) = n/(n,k).

Let d = n / (n,k). Then d(n,k) = (dn,dk) = n.

Let r = k / (n,k). Then nr = n(k / (n,k)) = k(n/ (n,k)) = dk.

Therefore, (dn,dk) = (dn, nr) = n and n = n(d,r), which holds if and only if (d,r) = 1. (QED)

**C**. Each subgroup **H** of **G** is of the form ‹
*a*^{d} ›, where d is a uniquely determined positive divisor of n.

__Proof__:

Let **H** be a subgroup of **G** and c be the order of **H**. Then, by Lagrange's theorem for finite groups, c divides n.

From part **B** we have that **o**(*a*^{k}) = c if and only if k = r(n/c) and (r,c) = 1, which is just to state that ‹ *a*^{k} › = **H** if and only if k = r(n/c) and (r,c) = 1

(1,c) = 1 and so if k = n/c then ‹ *a*^{k} › = **H** and n/(n/c) = c, so k obviously divides n.

Now, suppose ‹ *a*^{k} › = **H** and k divides n and k ≠ n/c.

This means that k = r(n/c) and (r,c) = 1 and r ≠ 1 and so r does not divide c. k, however, does divide n and so there exists an integer m such that km = n and k = n/m.

k = n/m = r(n/c). so 1/m = r/c and m = c/r contradicting the fact that r does not divide c.

Therefore k = n/c is the only divisor of n such that ‹ *a*^{k} › = **H**. (QED)

**D**. For each divisor d of n, the group **G** has exactly one subgroup **H**_{d} of order d and the number of elements of order d is equal to the number of positive integers less than d that are relatively prime to d and all of these elements lie in **H**_{d}.

__Proof__:

It follows from **B** that if **G** is a cyclic group of order n, then for each divisor d of n each and each 1 ≤ k < n,

*a*^{k} generates a subgroup **H** of **G** if and only if k = r(n/d) and r and d are relatively prime. By Lagrange's theorem, the order of each subgroup **H** of **G** must divide the order of **G**, i.e., n. Therefore, there are precisely φ(d) generators for every subgroup **H** of **G** of order d., where φ(d) is the number of positive integers less than or equal to d that are relatively prime to d. Now, suppose there is another subgroup ‹
*b* › of **G** of order d. Being cyclic, ‹
*b* › has an element of order d, which is necessarily of the form *b*^{r(n/d)} where (r,d) = 1. This element, however, is also in **H**_{d} = ‹ *a*^{(n/d)} › and is a generator of this subgroup. Therefore, ‹
*b* › = **H**_{d}. (QED)

**E**. For each positive integer n and each divisor d of n, Σ φ(d) = n, where φ is the Euler Phi Function.

__Proof__: Let **G** = ‹
*a* › be a cyclic group of order n. For each divisor d of n, let **G**_{d} = { *b* ε **G** | **o**(*b*) = d }. Again, by Lagrange's theorem, the order of each element must divide the order of **G**, and so each element *b* of **G** is in **G**_{d} for some divisor d of n. Furthermore, the set of all **G**_{d} forms a partition of the group **G**, given that the order of any element of **G** is unique. Therefore, the sum of the cardinalities of each **G**_{d} is n, but from **B**, for each d, the cardinality of **G**_{d} is φ(d). This just means that for each divisor d of n, Σ φ(d) = n. (QED)