# External direct products

1. Oct 6, 2004

### williamaholm

greetings, I'm confused on how to perform external direct products. I'm not sure how to symbolize this operation so I'll use +.
My text explains it as: G1+G2+...+Gn ={(g1,g2,...gn)}|giEGi}, where (g1,g2,...,gn)(g'1,g'2,...,g'n) is defined as (g1g'1, g2g'2,..., gng'n). It then gives this as an example: U(8)+U(10) = {(1,1),(1,3),(1,7),(1,9),(3,1),(3,3),(3,7),(3,9),(5,1),(5,3),(5,7),(5,9),(7,1),(7,3),(7,7),(7,9)} . Since U(8)={1,3,5,7} and U(10)={1,3,7,9} doesn't this example have the form of (g1g'1, g1g'2, g1g'3, g1g'4, g2g'1, g2g'2, g2g'3.... and so forth until we cycled through it all)

2. Oct 6, 2004

### matt grime

GxH is the set of all ordered pairs (g,h) where g is in G and h in H. Which is what you've written. And?
It is a group under the operation (g,h)*(g',h')=(g*g',h*h') where we compose elements of G in the obvious way, and elements of H similarly.

I can't decipher what you think is wrong.

3. Oct 6, 2004

### williamaholm

still confused

"doesn't this example have the form of (g1g'1, g1g'2, g1g'3, g1g'4, g2g'1, g2g'2, g2g'3...."
g1 is being distributed among all of g' s elements and then g2 gets distributed among all of g' s elements and this continues for all g elements. rather than g1g'1, g2g'2, g3g'3, where each element in g is associated with only one element in g'. Does this help you see where I'm confused?

4. Oct 7, 2004

### matt grime

No. Clearly the list you give with numbers in has more commas and different brackets. The list with g1 etc is a completely different format.

g1 distributed amongst the g's? what g's?

Read the definition of what the elements in the direct product are (ORDERED PAIRS, you have no ordered pairs mentioned in post #3.)

this bit here:
(g1g'1, g2g'2,..., gng'n)

refers to how you calculate the composition of two elements in the direct product.