Direct Products


Definition. Let G and H be groups. The direct product $G
   \times H$ of G and H is the set of all ordered pairs $\{(g,h) \mid g
   \in G, h \in H\}$ with the operation

$$(g_1, h_1)\cdot (g_2, h_2) = (g_1g_2, h_1h_2).$$

Remarks. 1. In the definition, I've assumed that G and H are using multiplication notation. In general, the notation you use in $G \times H$ depends on the notation in the factors. Examples:

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & G & & H & & \hbox{\vbox{\hbox{Product}\vskip2pt\hbox{($G \times H$)}}} & & \hbox{\vbox{\hbox{Identity}\vskip2pt\hbox{($G \times H$)}}} & & \hbox{\vbox{\hbox{Inverse}\vskip2pt\hbox{($G \times H$)}}} & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $g_1\cdot g_2$ & & $h_1\cdot h_2$ & & $(g_1,h_1)(g_2,h_2) = (g_1g_2,h_1h_2)$ & & $(1,1)$ & & $(g,h)^{-1} = (g^{-1},h^{-1})$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $g_1 + g_2$ & & $h_1 + h_2$ & & $(g_1,h_1) + (g_2,h_2) = (g_1 + g_2, h_1 + h_2)$ & & $(0,0)$ & & $-(g,h) = (-g,-h)$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $g_1\cdot g_2$ & & $h_1 + h_2$ & & $(g_1,h_1)(g_2,h_2) = (g_1g_2, h_1 + h_2)$ & & $(1,0)$ & & $(g,h)^{-1} = (g^{-1}, -h)$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

2. You can construct products of more than two groups in the same way. For example, if $G_1$ , $G_2$ , and $G_3$ are groups, then

$$G_1 \times G_2 \times G_3 = \{(x, y, z) \mid x \in G_1, y \in G_2, z \in G_3\}.$$

Just as with the two-factor product, you multiply elements componentwise.


Example. ( A product of cyclic groups which is cyclic) Since $\integer_2 = \{0, 1\}$ and $\integer_3 = \{0, 1, 2\}$ ,

$$\integer_2 \times \integer_3 = \{(0,0), (0,1), (0,2), (1,0), (1,1), (1,2)\}.$$

If you take successive multiples of $(1,1)$ , you get

$$(1,1), \quad (0,2), \quad (1,0), \quad (0,1), \quad (1,2), \quad (0,0).$$

Since you can get the whole group by taking multiples of $(1,1)$ , it follows that $\integer_2 \times
   \integer_3$ is actually cyclic of order 6 --- the same as $\integer_6$ .


Example. ( A product of cyclic groups which is not cyclic) Since $\integer_2 = \{0, 1\}$ ,

$$\integer_2 \times \integer_2 = \{(0,0), (1,0), (0,1), (1,1)\}.$$

Here's the operation table:

$$\vbox{\offinterlineskip \halign{ & \vrule # & \strut \hfil \quad # \quad \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & $(0,0)$ & & $(1,0)$ & & $(0,1)$ & & $(1,1)$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $(0,0)$ & & $(0,0)$ & & $(1,0)$ & & $(0,1)$ & & $(1,1)$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $(1,0)$ & & $(1,0)$ & & $(0,0)$ & & $(1,1)$ & & $(0,1)$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $(0,1)$ & & $(0,1)$ & & $(1,1)$ & & $(0,0)$ & & $(1,0)$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $(1,1)$ & & $(1,1)$ & & $(0,1)$ & & $(1,0)$ & & $(0,0)$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

Note that this is not the same group as $\integer_4$ . Both groups have 4 elements, but $\integer_4$ is cyclic of order 4. In $\integer_2 \times \integer_2$ , all the elements have order 2, so no element generates the group.

$\integer_2 \times \integer_2$ is the same as the Klein 4-group V, which has the following operation table:

$$\vbox{\offinterlineskip \halign{ & \vrule # & \strut \hfil \quad # \quad \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & 1 & & a & & b & & c & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 1 & & 1 & & a & & b & & c & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & a & & a & & 1 & & c & & b & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & b & & b & & c & & 1 & & a & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & c & & c & & b & & a & & 1 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

Compare the diagonal entries of the two tables, for instance.


Example. ( The order of a product) If G and H are finite, then $|G \times H| = |G||H|$ . For example, $|\integer_5 \times
   \integer_6| = 30$ .


Lemma. The product of abelian groups is abelian: If G and H are abelian, so is $G \times H$ .

Proof. Suppose G and H are abelian. Let $(g,h), (g',h') \in G \times H$ , where $g,
   g' \in G$ and $h, h' \in H$ . I have

$$\matrix{(g,h)(g',h') & = & (gg',hh') & \hbox{(Definition of multiplication in a product)} \cr & = & (g'g,h'h) & \hbox{(G and H are abelian)} \cr & = & (g',h')(g,h) & \hbox{(Definition of multiplication in a product)} \cr}$$

This proves that $G \times H$ is abelian.

On the other hand, if either G or H is not abelian, then $G \times H$ is not abelian. Suppose, for instance, that G is not abelian. This means that there are elements $g_1, g_2 \in G$ such that

$$g_1g_2 \ne g_2g_1.$$

Then

$$(g_1,1)(g_2,1) = (g_1g_2,1), \quad\hbox{while}\quad (g_2,1)(g_1,1) = (g_2g_1,1).$$

Since $(g_1g_2,1) \ne (g_2g_1,1)$ , it follows that $(g_1,1)(g_2,1) \ne (g_2,1)(g_1,1)$ , so $G \times H$ is not abelian.

A similar argument works if H is not abelian.


Example. ( A product of an abelian and a nonabelian group) The next two tables comprise the multiplication table for $\integer_2
   \times D_3$ . (Recall that $D_3$ is the group of symmetries of an equilateral triangle.) The number of elements is

$$|\integer_2 \times D_3| = |\integer_2|\cdot |D_3| = 2\cdot 6 = 12.$$

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $\cdot$ & & (0, id) & & (0, $r_1$) & & (0, $r_2$) & & (0, $m_1$) & & (0, $m_2$) & & (0, $m_3$) & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & (0, id) & & (0, id) & & (0, $r_1$) & & (0, $r_2$) & & (0, $m_1$) & & (0, $m_2$) & & (0, $m_3$) & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & (0, $r_1$) & & (0, $r_1$) & & (0, $r_2$) & & (0, id) & & (0, $m_3$) & & (0, $m_1$) & & (0, $m_2$) & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & (0, $r_2$) & & (0, $r_2$) & & (0, id) & & (0, id) & & (0, $m_2$) & & (0, $m_3$) & & (0, $m_1$) & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & (0, $m_1$) & & (0, $m_1$) & & (0, $m_2$) & & (0, $m_3$) & & (0, id) & & (0, $r_1$) & & (0, $r_2$) & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & (0, $m_2$) & & (0, $m_2$) & & (0, $m_3$) & & (0, $m_1$) & & (0, $r_2$) & & (0, id) & & (0, $r_1$) & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & (0, $m_3$) & & (0, $m_3$) & & (0, $m_1$) & & (0, $m_2$) & & (0, $r_1$) & & (0, $r_2$) & & (0, id) & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & (1, id) & & (1, id) & & (1, $r_1$) & & (1, $r_2$) & & (1, $m_1$) & & (1, $m_2$) & & (1, $m_3$) & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & (1, $r_1$) & & (1, $r_1$) & & (1, $r_2$) & & (1, id) & & (1, $m_3$) & & (1, $m_1$) & & (1, $m_2$) & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & (1, $r_2$) & & (1, $r_2$) & & (1, id) & & (1, id) & & (1, $m_2$) & & (1, $m_3$) & & (1, $m_1$) & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & (1, $m_1$) & & (1, $m_1$) & & (1, $m_2$) & & (1, $m_3$) & & (1, id) & & (1, $r_1$) & & (1, $r_2$) & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & (1, $m_2$) & & (1, $m_2$) & & (1, $m_3$) & & (1, $m_1$) & & (1, $r_2$) & & (1, id) & & (1, $r_1$) & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & (1, $m_3$) & & (1, $m_3$) & & (1, $m_1$) & & (1, $m_2$) & & (1, $r_1$) & & (1, $r_2$) & & (1, id) & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $\cdot$ & & (1, id) & & (1, $r_1$) & & (1, $r_2$) & & (1, $m_1$) & & (1, $m_2$) & & (1, $m_3$) & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & (0, id) & & (1, id) & & (1, $r_1$) & & (1, $r_2$) & & (1, $m_1$) & & (1, $m_2$) & & (1, $m_3$) & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & (0, $r_1$) & & (1, $r_1$) & & (1, $r_2$) & & (1, id) & & (1, $m_3$) & & (1, $m_1$) & & (1, $m_2$) & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & (0, $r_2$) & & (1, $r_2$) & & (1, id) & & (1, id) & & (1, $m_2$) & & (1, $m_3$) & & (1, $m_1$) & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & (0, $m_1$) & & (1, $m_1$) & & (1, $m_2$) & & (1, $m_3$) & & (1, id) & & (1, $r_1$) & & (1, $r_2$) & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & (0, $m_2$) & & (1, $m_2$) & & (1, $m_3$) & & (1, $m_1$) & & (1, $r_2$) & & (1, id) & & (1, $r_1$) & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & (0, $m_3$) & & (1, $m_3$) & & (1, $m_1$) & & (1, $m_2$) & & (1, $r_1$) & & (1, $r_2$) & & (1, id) & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & (1, id) & & (0, id) & & (0, $r_1$) & & (0, $r_2$) & & (0, $m_1$) & & (0, $m_2$) & & (0, $m_3$) & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & (1, $r_1$) & & (0, $r_1$) & & (0, $r_2$) & & (0, id) & & (0, $m_3$) & & (0, $m_1$) & & (0, $m_2$) & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & (1, $r_2$) & & (0, $r_2$) & & (0, id) & & (0, id) & & (0, $m_2$) & & (0, $m_3$) & & (0, $m_1$) & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & (1, $m_1$) & & (0, $m_1$) & & (0, $m_2$) & & (0, $m_3$) & & (0, id) & & (0, $r_1$) & & (0, $r_2$) & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & (1, $m_2$) & & (0, $m_2$) & & (0, $m_3$) & & (0, $m_1$) & & (0, $r_2$) & & (0, id) & & (0, $r_1$) & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & (1, $m_3$) & & (0, $m_3$) & & (0, $m_1$) & & (0, $m_2$) & & (0, $r_1$) & & (0, $r_2$) & & (0, id) & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The operation in $\integer_2$ is addition mod 2, while the operation in $D_3$ is written using multiplicative notation. When you multiply two pairs, you add in $\integer_2$ in the first component and multiply in $D_3$ in the second component:

$$(1,r_2)(1,m_2) = (1 + 1, r_2\cdot m_2) = (0,m_3).$$

The identity is $(0,\hbox{id})$ , since 0 is the identity in $\integer_2$ , while id is the identity in $D_3$ .

$\integer_2 \times D_3$ is not abelian, since $D_3$ is not abelian. A particular example:

$$(1,m_2)(0,r_2) = (1,m_1), \quad\hbox{but}\quad (0,r_2)(1,m_2) = (1,m_3).\quad\halmos$$


Example. ( Using products to construct groups) You can use products to construct 3 different abelian groups of order 8.

The groups $\integer_2 \times
   \integer_2 \times \integer_2$ , $\integer_4 \times \integer_2$ , and $\integer_8$ are abelian, since each is a product of abelian groups. $\integer_8$ is cyclic of order 8, $\integer_4 \times
   \integer_2$ has an element of order 4 but is not cyclic, and $\integer_2 \times \integer_2 \times \integer_2$ has only elements of order 2. It follows that these groups are distinct.

The group $D_4$ 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.

$D_4$ or Q are not that same as $\integer_2 \times \integer_2 \times
   \integer_2$ , $\integer_4 \times \integer_2$ , or $\integer_8$ , since $\integer_2 \times \integer_2 \times
   \integer_2$ , $\integer_4 \times \integer_2$ , and $\integer_8$ are abelian while $D_4$ or Q are not.

Finally, $D_4$ is not the same as Q. $D_4$ has 5 elements of order 2: The four reflections and rotation through $180^\circ$ . 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 $[m,n]$ of m and n is the smallest positive integer divisible by m and n.

Remark. Since $mn$ 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, $[18,30] = 90$ .

Lemma. If s is a common multiple of m and n, then $[m,n] \mid s$ .

Proof. By the Division Algorithm,

$$s = q\cdot [m,n] + r, \quad\hbox{where}\quad 0 \le r < [m,n].$$

Thus, $r = s - q\cdot [m,n]$ . Since $m
   \mid s$ and $m \mid [m,n]$ , I have $m \mid r$ . Since $n \mid s$ and $n \mid [m,n]$ , I have $n \mid r$ . Therefore, r is a common multiple of m and n. Since it's also less than the least common multiple $[m,n]$ , it can't be positive. Therefore, $r = 0$ , and $s = q\cdot [m,n]$ , i.e. $[m,n] \mid s$ .

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

$$mn = (m,n)[m,n].$$

Proof. I'll prove that each side is greater than or equal to the other side.

Note that $\dfrac{m}{(m,n)}$ and $\dfrac{n}{(m,n)}$ are integers. Thus,

$$\dfrac{mn}{(m,n)} = m\cdot \dfrac{n}{(m,n)} = \dfrac{m}{(m,n)}\cdot n$$

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,

$$\dfrac{mn}{(m,n)} \ge [m,n], \quad\hbox{and}\quad mn \ge (m,n)[m,n].$$

Next, $[m,n]$ is a multiple of n, so $[m,n] = sn$ for some s. Then

$$\dfrac{mn}{[m,n]} = \dfrac{mn}{sn} = \dfrac{m}{s} \mid m.$$

(Why is $\dfrac{mn}{[m,n]}$ an integer? Well, $mn$ is a common multiple of m and n, so by the previous lemma $[m,n] \mid mn$ .)

Similarly, $[m,n]$ is a multiple of m, so $[m,n] = tm$ for some t. Then

$$\dfrac{mn}{[m,n]} = \dfrac{mn}{tm} = \dfrac{n}{t} \mid n.$$

In other words, $\dfrac{mn}{[m,n]}$ is a common divisor of m and n. Therefore, it must be less than the greatest common divisor:

$$\dfrac{mn}{[m,n]} \le (m,n), \quad\hbox{and}\quad mn \le (m,n)[m,n].$$

The two inequalities I've proved show that $mn = (m,n)[m,n]$ .


Example. ( The order of an element in a product) $(54, 72) =
   18$ , $[54, 72] = 216$ , and

$$(54, 72)[54, 72] = 18\cdot 216 = 3888 = 54\cdot 72.\quad\halmos$$


Proposition. The element $(1,1)$ has order $[m,n]$ in $\integer_m \times
   \integer_n$ .

Proof.

$$[m,n](1,1) = ([m,n], [m,n]).$$

The first component is 0, since it's divisible by m; the second component is 0, since it's divisible by n. Hence, $[m,n](1,1) = (0,0)$ .

Next, I must show that $[m,n]$ is the smallest positive multiple of $(1,1)$ which equals the identity. Suppose $k(1,1) = (0,0)$ , so $(k,k) = (0,0)$ . Consider the first components. $k = 0$ in $\integer_m$ means that $m \mid k$ ; likewise, the second components show that $n \mid k$ . Since k is a common multiple of m and n, it must be greater than or equal to the least common multiple $[m,n]$ : that is, $k \ge [m,n]$ . This proves that $[m,n]$ is the order of $(1,1)$ .


Example. ( The order of (1,1)) In $\integer_4 \times
   \integer_6$ , the element $(1,1)$ has order $[4,6] = 12$ .

On the other hand, the element $(1,1)
   \in \integer_5 \times \integer_6$ has order 30. Since $\integer_5 \times \integer_6$ has order 30, the group is cyclic; in fact, $\integer_5 \times \integer_6 \approx
   \integer_{30}$ .


Remark. More generally, consider $(x_1, \ldots, x_n) \in G_1 \times \ldots
   \times G_n$ , and suppose $x_i$ has order $r_i$ in $G_i$ . (The $G_i$ 's need not be cyclic.) Then $(x_1, \ldots, x_n)$ has order $[r_1, \ldots, r_n]$ .

Corollary. $\integer_m \times \integer_n$ is cyclic of order $mn$ if and only if $(m,n) = 1$ .

Proof. If $(m,n) = 1$ , then $[m,n] = mn$ . Thus, the order of $(1,1)$ is $[m,n] = mn$ . But $\integer_m
   \times \integer_n$ has order $mn$ , so $(1,1)$ generates the group. Hence, $\integer_m \times \integer_n$ is cyclic.

Suppose on the other hand that $(m,n)
   \ne 1$ . Since $(m,n)[m,n] = mn$ , it follows that $[m,n] \ne mn$ . Since $mn$ is a common multiple of m and n and since $[m,n]$ is the least common multiple, it follows that $[m,n] < mn$ .

Now consider an element $(a,b) \in
   \integer_m \times \integer_n$ . Let p be the order of a in $\integer_m$ and let q be the order of b in $\integer_n$ .

Since $p \mid m \mid [m,n]$ , I may write $pj = [m,n]$ for some j. Since $q \mid n \mid [m,n]$ , I may write $qk = [m,n]$ for some k. Then

$$[m,n](a,b) = ([m,n]a, [m,n]b) = (j(pa), k(qb)) = (j\cdot 0, k\cdot 0) = (0,0).$$

Hence, the order of $(a,b)$ is less than or equal to $[m,n]$ . But $[m,n] < mn$ , so the order of $(a,b)$ is less than (and not equal to) $mn$ .

Since $(a,b)$ was an arbitrary element of $\integer_m \times \integer_n$ , it follows that no element of $\integer_m \times \integer_n$ has order $mn$ . Therefore, $\integer_m \times \integer_n$ can't be cyclic of order $mn$ , since a generator would have order $mn$ .

Remark. More generally, if $m_1$ , ..., $m_k$ are pairwise relatively prime, then $\integer_{m_1} \times \ldots \times
   \integer_{m_k}$ is cyclic of order $m_1\cdots m_k$ .


Example. ( Orders of elements in products) $\integer_5
   \times \integer_6$ is cyclic of order 30. This means that as a group, it's "the same as" $\integer_{30}$ . However, $\integer_2
   \times \integer_2$ is not cyclic of order 4.

Using an earlier remark, the element $(2, 4, 4)$ in $\integer_4 \times \integer_{12} \times
   \integer_6$ has order $[2, 3, 3] = 6$ --- because 2 has order 2 in $\integer_4$ , 4 has order 3 in $\integer_{12}$ , and 4 has order 3 in $\integer_6$ .


Example. ( A product of cyclic groups which is not cyclic) $\integer_2 \times \integer_4$ is a group of order $2\cdot 4 = 8$ . However, it is not cyclic of order 8, since $(2,4) = 2 \ne 1$ . In fact, if $(a,b) \in \integer_2 \times
   \integer_4$ , then

$$4(a,b) = (4a, 4b) = (0,0).$$

Thus, every element of $\integer_2
   \times \integer_4$ has order less than or equal to 4. In particular, there can be no elements of order 8, i.e. no cyclic generators.


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