We have introduced the order of an element; now we develop its theory more fully and use it to classify cyclic groups.
Proposition 2.6.1 (Powers and order). Let .
If has finite order , then , and . The elements are all distinct, and ; so .
If has infinite order, then the powers ,, are pairwise distinct; so is infinite.
In either case, .
Proof.(a)() If , then .() Write with . Then . By minimality of ,, so . Next, . If for , then would divide , with , which is impossible; so are distinct. Finally, every power equals , where is the remainder of on division by ; so .
(b) If with , then with , contradicting infinite order. ∎
Theorem 2.6.2 (Structure of cyclic groups). Let .
If , then ,, is an isomorphism, so .
If , then , these elements are distinct, and ,(), is an isomorphism, so .
Thus every cyclic group is isomorphic to or to for some .
Proof. The map ,, is a homomorphism with image (Example 2.5.7), so it is surjective.
(a) If , the powers of are pairwise distinct by Proposition 2.6.1, so is injective, hence an isomorphism .
(b) If , Proposition 2.6.1 shows that with these elements distinct, so ,, is a bijection. It is a homomorphism: differs from by a multiple of , so by Proposition 2.6.1, .∎
Note. In case (b), by Proposition 2.6.1, and is the map that induces on . In the next chapter this becomes an instance of the First Isomorphism Theorem (Theorem 3.3.3): .
Example 2.6.3 (Cyclic groups in disguise).
has order , so via .
The rotation subgroup of is isomorphic to , since has order .
In , the element satisfies if and only if and , that is, . So has order , the subgroup it generates has elements, and . By Theorem 2.6.2, is an isomorphism . On the other hand, , since the former has no element of order 4.
Proposition 2.6.4 (Order of powers). Let , and let . Then .
Proof. Let , and write and . Then : if divides both and , then divides both and , so and .
First, , so is finite and divides by Proposition 2.6.1.
Conversely, let . Then , so , that is, , and so . Since , Euclid’s lemma (Corollary 2.4.8) gives . As and are positive and divide each other, .∎
Corollary 2.6.5 (Generators of cyclic groups). In a cyclic group of order , the element is a generator iff .
Proof. Since and both are finite, if and only if , that is, (Proposition 2.6.1). By Proposition 2.6.4, this happens if and only if .∎
Example 2.6.6. In , the generators are (the four elements coprime to 12). The element has order .
Proposition 2.6.7 (Subgroups of cyclic groups). Every subgroup of a cyclic group is cyclic. More precisely, if has order , then the subgroups of are exactly the subgroups for the positive divisors of ; these are distinct, and . So has exactly one subgroup of each order dividing .
Proof. Let . If , then is cyclic. Otherwise contains some , and with it , so contains a positive power of . Let be the smallest positive integer with . We claim . Clearly . Conversely, take , and write with . Then . By minimality of ,, so . So every subgroup of is cyclic.
Now suppose , so that by Proposition 2.6.1. If and is as above, then , and the same division argument, applied to , shows that . If , then , and divides . So every subgroup of is for some positive divisor of . For such , Propositions 2.6.1 and 2.6.4 give . Different divisors give different orders , so the subgroups are distinct, and as runs over the divisors of , so does .∎
Question 2.6.8.
Find all subgroups of and draw the subgroup lattice.
Prove that has no proper nontrivial subgroups for prime.
How many generators does have?
Prove that for .
Prove: if is a group in which every element satisfies , then is abelian.
Show solution
(a) By Proposition 2.6.7, the subgroups are for the divisors of 12, namely 1, 2, 3, 4, 6, 12: ,,,,, and . Moreover, exactly when divides : if , then ; conversely, the elements of are the multiples of in (because ), so forces .
Lattice (larger subgroups higher, an edge for each inclusion with no subgroup in between):
Note that , since is not a multiple of .
(b) By Proposition 2.6.7, the subgroups of correspond to the positive divisors of , which are and . They are and .
(c) The element is , the th power of the generator in additive notation. By Corollary 2.6.5, it generates iff . The count is , Euler’s totient.
(d) An automorphism of is determined by , since . Its image must be all of , so is a generator, i.e., by (c), i.e., . Conversely, for , let . It is a homomorphism, since and differ by a multiple of ; and if is the inverse of in , then and are multiplication by , the identity, so is an automorphism. The map satisfies , so it is a homomorphism . It is injective, since , and surjective by the first part (). So it is an isomorphism.
(e) For any : (using , since ). So is abelian.
Question 2.6.9.
In a group , if and , what are the possible orders of ?
Prove: if and with and , then .
Show solution
(a) By Proposition 2.6.1, divides both 12 and 8, so by Proposition 2.4.6. Possible orders: 1, 2, 4. All three occur, for instance for ,, and in .
(b) Let . Since and commute, , so is finite and .
Now (since ), so . Since , Euclid’s lemma gives . Similarly, . Writing , we have , so by Euclid’s lemma again, and . Thus .
Question 2.6.10.
Prove that the order of is the least common multiple of and (when both are finite), and find the order of in .
Prove that if and only if .
Find all groups of order 4 up to isomorphism.
Prove that for any groups .
Show solution
(a) Let and . Since , Proposition 2.6.1 gives if and only if and , that is, . This intersection is a subgroup of (Proposition 2.4.4) containing , so by Theorem 2.4.5 it is , where is its smallest positive element, the least common multiple . So . For : in and in (Proposition 2.6.4), so .
(b)() Assume . Let and be generators of and (for instance, the remainders of ), of orders and . By (a), has order . A common multiple of and satisfies , hence by Euclid’s lemma, so ; thus . So , a group of order , contains an element of order ; by Proposition 2.6.1 the cyclic subgroup it generates has elements, so it is the whole group. Hence the group is cyclic of order , and by Theorem 2.6.2.
() If , then for any , we have , since and . So every element has order dividing , and no element has order . But has an element of order , so the two groups are not isomorphic (Proposition 2.5.9).
(c) Let have order 4. A non-identity element cannot have order greater than , since then would be five distinct elements (Proposition 2.6.1). Nor can it have order : then for some , and would have to be one of these four elements, but gives , gives , gives , and gives . So every non-identity element has order 2 or 4.
Case 1: has an element of order 4. Then .
Case 2: Every non-identity element has order 2. Then with . What is ? It’s not (else ), not (else ), and not (else ). So , and by the same argument with and exchanged, . Likewise and . So the multiplication table of is determined, and it matches that of under ,,,. So , the Klein four-group.
These two groups are not isomorphic, so up to isomorphism there are exactly two groups of order 4.
(d) Define by . It is a bijection, its inverse being . It is a homomorphism: .