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July 5th, 2014, 12:03 AM   #1
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direct product

What's a direct product of the cyclic groups of order m and n?
I know the cardinality is m*n but I really cannot understand how it works.
Can the generator of these two cycles group be a permutation? And if yes, this means that one is a permutation of length a and the other of length b. So what is the result of such a product?

Last edited by skipjack; July 5th, 2014 at 01:35 AM.
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July 8th, 2014, 10:22 AM   #2
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The usual way to understand this, is to write this group ADDITIVELY (as $\Bbb Z_m \times \Bbb Z_n$).

So the operation is +, where:

$(a,b) + (c,d) = (a+c,b+d)$

and the first coordinate is taken mod $m$, the second mod $n$.

The identity is $(0,0)$, and the inverse of $(a,b)$ is:

$-(a,b) = (-a,-b) = (m-a,n-b)$, and "powers" in this group are "multiples":

$k(a,b) = (a,b) + \cdots + (a,b)$ ($k$ times).

Let's look at two examples, which will indicate how these things typically work.

First we'll look at $\Bbb Z_2 \times \Bbb Z_3$ which has 6 elements:


This group is cyclic, with (1,1) as a generator:

(1,1) + (1,1) = (0,2)
(0,2) + (1,1) = (1,0)
(1,0) + (1,1) = (0,1)
(0,1) + (1,1) = (1,2)
(1,2) + (1,1) = (0,0)

(the element (1,2) is also a generator, (1,0) has order 2, (0,1) and (0,2) have order 3).

Is this always true? Unfortunately, no, it happens here because lcm(2,3) = 6, which is the order of the group. Let's look at $\Bbb Z_4 \times \Bbb Z_6$ where lcm(4,6) = 12.

Here, we see that $(1,1)$ is NOT a generator, because:

$12(1,1) = 12(1,0) + 12(0,1) = 3(4(1,0)) + 2(6(0,1)) = 3(0,0) + 2(0,0) = 5(0,0) = (0,0)$.

(Note that these are almost like vector spaces, with "scalars" being integers).

It turns out no element of $\Bbb Z_4 \times \Bbb Z_6$ has order greater than 12, since:

$12(a,b) = 12(a,0) + 12(0,b) = a(12(1,0)) + b(12(0,1)) = a(0,0) + b(0,0) = (0,0)$.

If one thinks of these as "cycles", say the disjoint pair of cycles:

$(a_1\ a_2\ \dots\ a_m)(b_1\ b_2\ \dots\ b_n) \in S_{m+n}$

you can see that the only time the $a$'s and $b$'s SIMULTANEOUSLY return to their starting places is at a power which is a common multiple of $m$ and $n$. Thus this par of cycles has order lcm(m,n).
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