* Definition.* Let G and H be groups. The * direct product* of G and H is the set
of all ordered pairs with the
operation

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* 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.

Since ,

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

Then

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 .

* Proof.*

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.

Copyright 2018 by Bruce Ikenaga