I have this exercise and there are various parts, I write down here only the first part.

Without the comprehension of this part I can't do anything in the remaining parts.

G and H are two groups.

Let GxH = {(x,y) : x ∈ G and y ∈ H}

The operation GxH consists of multiplying corresponding components:

(x,y) ∙ (x',y') = (xx', yy')

If G and H are donoted additively, it is customary to also denote GxH additively:

(x,y) + (x',y') = (x+x', y+y')

Part 1:

Prove that GxH is a group by checking the three group axioms:

(G1)

(x

_{1},y

_{1})[(x

_{2},y

_{2})(x

_{3},y

_{3})] =

[(x

_{1},y

_{1})(x

_{2},y

_{2})](x

_{3},y

_{3}) =

(G2)

Let e

_{G}be the identity element of G, and e

_{H}the identity element of H.

Find and check the identity element of GxH?

(G3)

For each (a,b) ∈ GxH, finf and check the inverse of (a,b).

Part2:

List the elements of Z

_{2}xZ

_{3}, and write its operation table.

(Note: There are six elements, each of which is an ordered pair. The notation is additive).

development:

The are some other parts, but I stop here. I want to comprehend these ones.

I have tried somethings like these, but without succcess..:

For part1:

(G1)

(x

_{1}, y

_{1})[(x

_{2}, y

_{2})(x

_{3}, y

_{3})] =

(x

_{1}, y

_{1})(x

_{2}x

_{3}, y

_{2}y

_{3}) =

(x

_{1}x

_{2}x

_{3}, y

_{1}y

_{2}y

_{3})

[(x

_{1}, y

_{1})(x

_{2}, y

_{2})](x

_{3}, y

_{3}) =

(x

_{1}x

_{2}, y

_{1}y

_{2})(x

_{3}, y

_{3}) =

(x

_{1}x

_{2}x

_{3}, y

_{1}y

_{2}y

_{3})

(G2)

If I am right, We have a single identity element for each group so:

(x, e

_{H}) ∙ (x', e

_{H}) = (xx', e

_{H})

(e

_{G}, y) ∙ (e

_{G}, y') = (e

_{G}, yy')

(xx', e

_{H}) ∙ (e

_{G}, yy') = (xx'e

_{G}, e

_{H}yy')

(G3)

I have denoted (a,b)' as the inverse of (a,b):

(a, b) ∙ (a, b)' = (xx'e

_{G}, e

_{H}yy')

For Part2:

Z

_{2}xZ

_{3}= {(0,0), (0,1), (0,2), (1,0), (1,1), (1,2)}

so the additive table is:

+_0_1_2

0_0_1_2

1_1_2_3

please can you help me? many thanks!