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)