Amber Bowling

My group is the Direct Product Group \$DP_2_,_2_,_2\$ (which basically means, \$\mathbb{Z}_{2} X \mathbb{Z}_{2} X \mathbb{Z}_{2}\$) under component-wise addition: (a,b,c) + (d,e,f) = (a+d, b+e, c+f).

Associativity:

Is [(a,b,c) + (d,e,f)] + (g,h,i) = (a,b,c) + [(d,e,f) + (g,h,i)] ?

Let [(a,b,c) + (d,e,f)] + (g,h,i) be written as [(a+d), (b+e), (c+f)] + (g,h,i)]

This in turn is equal to [(a+d) +g,(b+e) + f, (c+f) + i]

We know that (a+d) + g is computed in \$\mathbb{Z}_{2}\$ as are (b+e) + f and (c+f) + i,
thus we can use the assoiciativity property of \$\mathbb{Z}_{2}\$ to find :

[a + (d+g), b (e+h), c + (f+i)] which is also equal to [(a,b,c) + [(d+g), (e+h), (f+i)]].

We can then go one step further to find that : (a,b,c) + [(d,e,f) + (g,h,i)]

Hence [(a,b,c) + (d,e,f)] + (g,h,i) = (a,b,c) + [(d,e,f) + (g,h,i)]. Therefore associativity holds.

Identity:

Is [e + (a,b,c) = (a,b,c)]

Let e = (0,0,0)

We then find that: [ (0,0,0) + (a,b,c) = (a,b,c)] so the identity holds. We know this because in \$\mathbb{Z}_{2}\$ the only options we have are 0 and 1. In this case 0 is what we need.

Inverse:

Is [ (a, b, c) + \$(a^{-1},b^{-1},c^{-1})\$ = (0, 0, 0)?

Let (a,b,c) = (0,1,0)

In \$\mathbb{Z}_{2}\$ what is this triples only option for an inverse?

We need [ (0, 1, 0) + \$(a^{-1},b^{-1},c^{-1})\$ = (0, 0, 0).

Which would be:

[ (0, 1, 0) + (0, 1, 0) = (0, 0, 0)

We can see from above that the inverse for (a,b,c) is simply (a,b,c). Thus inverse holds.

Since associativity, identity, and inverse hold we see that \$DP_2_,_2_,_2\$ is a group!!

Subgroups and Left Cosets of \$DP_2_,_2_,_2\$

Let (a,b,c) = (1,0,1) and (d,e,f) = (1,0,1).

Then (a,b,c) + (d,e,f) = (1+1, 0+0, 1+1) = (0,0,0)

Thus (0,0,0) is a subgroup.

The left cosets are as follows:

( ) + (0,0,0) = (0,0,0)

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

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

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

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

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

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

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

page revision: 32, last edited: 24 Nov 2009 23:28