Definition. Let G and H be groups. The direct product of G and H is the set of all ordered pairs with the operation
Remarks. 1. In the definition, I've assumed that G and H are using multiplication notation. In general, the notation you use in depends on the notation in the factors. Examples:
2. You can construct products of more than two groups in the same way. For example, if , , and are groups, then
Just as with the two-factor product, you multiply elements componentwise.
Example. ( A product of cyclic groups which is cyclic) Show that is cyclic.
Since and ,
If you take successive multiples of , you get
Since you can get the whole group by taking multiples of , it follows that is actually cyclic of order 6 --- the same as .
Example. ( A product of cyclic groups which is not cyclic) Show that is not cyclic.
Here's the operation table:
Note that this is not the same group as . Both groups have 4 elements, but is cyclic of order 4. In , all the elements have order 2, so no element generates the group.
is the same as the Klein 4-group V, which has the following operation table:
If G and H are finite, then . (This is true for sets G and H; it has nothing to do with G and H being groups.) For example, .
Lemma. The product of abelian groups is abelian: If G and H are abelian, so is .
Proof. Suppose G and H are abelian. Let , where and . I have
This proves that is abelian.
Remark. If either G or H is not abelian, then is not abelian. Suppose, for instance, that G is not abelian. This means that there are elements such that
Since , it follows that , so is not abelian.
A similar argument works if H is not abelian.
Example. ( A product of an abelian and a nonabelian group) Construct the multiplication table for . (Recall that is the group of symmetries of an equilateral triangle.) The number of elements is
Here's the multiplication table for :
The operation in is addition mod 2, while the operation in is written using multiplicative notation. When you multiply two pairs, you add in in the first component and multiply in in the second component:
The identity is , since 0 is the identity in , while id is the identity in .
is not abelian, since is not abelian. A particular example:
Example. ( Using products to construct groups) Use products to construct 3 different abelian groups of order 8. The groups , , and are abelian, since each is a product of abelian groups. is cyclic of order 8, has an element of order 4 but is not cyclic, and has only elements of order 2. It follows that these groups are distinct.
In fact, there are 5 distinct groups of order 8; the remaining two are nonabelian.
The group of symmetries of the square is a nonabelian group of order 8.
The fifth (and last) group of order 8 is the group Q of the quaternions.
or Q are not that same as , , or , since , , and are abelian while or Q are not.
Finally, is not the same as Q. has 5 elements of order 2: The four reflections and rotation through . Q has one element of order 2, namely -1.
I've shown that these five groups of order 8 are distinct; it takes considerably more work to show that these are the only groups of order 8.
Definition. Let m and n be positive integers. The least common multiple of m and n is the smallest positive integer divisible by m and n.
Remark. Since is divisible by m and n, the set of positive multiples of m and n is nonempty. Hence, it has a smallest element, by well-ordering. It follows that the least common multiple of two positive integers is always defined. For example, .
Lemma. If s is a common multiple of m and n, then .
Proof. By the Division Algorithm,
Thus, . Since and , I have . Since and , I have . Therefore, r is a common multiple of m and n. Since it's also less than the least common multiple , it can't be positive. Therefore, , and , i.e. .
Remark. The lemma shows that the least common multiple is not just "least" in terms of size. It's also "least" in the sense that it divides every other common multiple.
Theorem. Let m and n be positive integers. Then
Proof. I'll prove that each side is greater than or equal to the other side.
Note that and are integers. Thus,
This shows that is a multiple of m and a multiple of n. Therefore, it's a common multiple of m and n, so it must be greater than or equal to the least common multiple. Hence,
Next, is a multiple of n, so for some s. Then
(Why is an integer? Well, is a common multiple of m and n, so by the previous lemma .)
Similarly, is a multiple of m, so for some t. Then
In other words, is a common divisor of m and n. Therefore, it must be less than the greatest common divisor:
The two inequalities I've proved show that .
Example. Verify that if and .
, , and
Proposition. The element has order in .
The first component is 0, since it's divisible by m; the second component is 0, since it's divisible by n. Hence, .
Next, I must show that is the smallest positive multiple of which equals the identity. Suppose , so . Consider the first components. in means that ; likewise, the second components show that . Since k is a common multiple of m and n, it must be greater than or equal to the least common multiple : that is, . This proves that is the order of .
Example. Find the order of in . Find the order of .
The element has order .
On the other hand, the element has order . Since has order 30, the group is cyclic; in fact, .
Remark. More generally, consider , and suppose has order in . (The 's need not be cyclic.) Then has order .
Corollary. is cyclic of order if and only if .
Note: In the next proof, " " may mean either the ordered pair or the greatest common divisor of a and b. You'll have to read carefully and determine the meaning from the context.
Proof. If , then . Thus, the order of is . But has order , so generates the group. Hence, is cyclic.
Suppose on the other hand that . Since , it follows that . Since is a common multiple of m and n and since is the least common multiple, it follows that .
Now consider an element . Let p be the order of a in and let q be the order of b in .
Since , I may write for some j. Since , I may write for some k. Then
Hence, the order of is less than or equal to . But , so the order of is less than (and not equal to) .
Since was an arbitrary element of , it follows that no element of has order . Therefore, can't be cyclic of order , since a generator would have order .
Remark. More generally, if , ..., are pairwise relatively prime, then is cyclic of order .
Example. ( Orders of elements in products) Find the order of .
2 has order 2 in , 4 has order 3 in , and 4 has order 3 in . Hence, the order of is .
Example. ( A product of cyclic groups which is not cyclic) Prove directly that is not cyclic of order 8.
If , then
Thus, every element of has order less than or equal to 4. In particular, there can be no elements of order 8, i.e. no cyclic generators.
Bruce Ikenaga's Home Page
Copyright 2018 by Bruce Ikenaga