vertices of the unit cube under component addition-sub-2

This group has 8 elements:
$0=(0,0,0)$
$a=(1,0,0)$
$b=(0,1,0)$
$c=(1,1,0)$
$d=(0,0,1)$
$f=(1,0,1)$
$g=(0,1,1)$
$h=(1,1,1)$

A generic element of this set shall be represented as $x=(x_{1}, x_{2}, x_{3})$, where $x_{1}, x_{2}, x_{3} \in \mathbb{Z}_{2}$

Verification of group:
Table:

 + 0 a b c d f g h 0 0 a b c d f g h a a 0 c b f d h g b b c 0 a g h d f c c b a 0 h g f d d d f g h 0 a b c f f d h g a 0 c b g g h d f b c 0 a h h g f d c b a 0

Associativity:
$x, y, z \in \mathbb{Z}_{2} ^3$
Then,
$(x_{1}+y_{1})+z_{1}=x_{1}+(y_{1}+z_{1})$,
$(x_{2}+y_{2})+z_{2}=x_{2}+(y_{2}+z_{2})$, and
$(x_{3}+y_{3})+z_{3}=x_{3}+(y_{3}+z_{3})$ by associativity of $+_{2}$ in $\mathbb{Z}_{2}$.
So,
$(x+y)+z=x+(y+z)$]
Therefore, component addition is associative in $\mathbb{Z}_2^3$.

Identy:
As the table shows, 0 is the identity.

Inverse:
As the table shows, each element is its own inverse.

Order of group:
$|\mathbb{Z}_2^3|=8$

Order of elements:
Because every element is its own inverse,
$|a|=|b|=|c|=|d|=|f|=|g|=|h|=2$

Because 0 is the identity, it generates the trivial subgroup, {0}, so $|0|=1$

Commutivity
Let $x,y\in\mathbb{Z}_2^3$ be arbitrary.

$x=(j,k,l), y=(m,n,p)$ where $j,k,l,m,n,p\in\mathbb{z}_2$

$x+y=(j+m,k+n,l+p)$ by definition of component addition

$x+y=(m+j,n+k,p+l)$ by commutivity of addition in $\mathbb{z}_2$

$x+y=(m,n,p)+(j,k,l)$ by definition of component addition

$x+y=y+x$

Therefore, $\mathbb{Z}_2^3$ is abelian

Generators
Generating sets:
{a,b,d},{a,b,f},{a,b,g},{a,b,h},{a,c,d},{a,c,f},{a,c,g},{a,c,h},{a,d,g},{a,d,h},{a,f,g},{a,f,h},{b,c,d},{b,c,f},{b,c,g},{b,c,h},{b,d,f},{b,d,h},{b,f,g},{b,g,h},{c,d,f},{c,d,g},{c,f,h},{d,f,g},{d,f,h},{f,g,h}
Any set of three elements such that the plane in $\mathbb{R}^3$ defined by those three elements contains no other elements of $\mathbb{Z}_2^3$, except {c,f,g}. Also, any set of three non-identity elements where two do not sum to the third.

Subgroups

Order 2:

(all subgroups generated by a single element)
{e,a},{e,b},{e,c},{e,d},{e,f},{e,g},{e,h}

Order 4:

(all subgroups generated by two elements)
{e,a,b,c},{e,a,d,f},{e,a,g,h},{e,b,d,g},{e,b,f,h},{e,c,d,h},{e,c,f,g}

Cosets
The cosets of {e,a} are {e,a},{b,c},{d,f}, and {g,h}.
These cosets are the edges of the cube that are parellel to {e,a}.

page revision: 16, last edited: 06 Dec 2010 17:27