If G and H are nay two groups, then direct product is a new group, denoted by G
ID: 1891641 • Letter: I
Question
If G and H are nay two groups, then direct product is a new group, denoted by G times H and defined as follows G times H consists of all the ordere pairs (x,y). Where x is in G and y is in H.G times H= {(x,y): x is in G and y is in H}
The operation G times H consists of multiplying corresponding components (x,y)(x',y')=(xx',yy').
If G and H are denoted additively (x,y)+(x',y')= (x+x', y+y')
1.) Prove: (x1, y1) [(x2,y2)(x3,y3)]= (x1x2x3, y1y2y3) i think, is it corect?
[(x1, y1)(x2, y2)] (x3, y3) = (x1x2x3, y1y2y3) Is this correct?
2.) Let eg be the identity element for G and eh be the identity element for H.
G times H = ( , ) and check
3.) For each (a, b) in G times H the inverse of (a, b) is ( , ).
explain
SHOW ALL STEPS
Explanation / Answer
1. (x1, y1)[(x2,y2)(x3,y3)] = (x1,y1)(x2x3,y2y3) = (x1x2x3, y1y2y3) [(x1,y1)(x2,y2)](x3,y3) = (x1x2,y1y2)(x3,y3) = (x1x2x3,y1y2y3) 2. The identity element of G times H = (eg, eh). Because let (x,y) be an arbitrary element of GxH. Then (eg,eh)(x,y) = (eg*x, e^h*y) = (x,y) = (x*eg,y*eh) = (x,y)(eg,eh). 3. For each (a,b) in GxH, the inverse of (a,b) is (a' , b') where a' is the inverse of a in G, and b' is the inverse of b in H. Since (a,b)(a',b') = (aa',bb') = (eg,eh) = (a'a, b'b) = (a',b')(a,b)