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

This group has 8 elements:

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:

+ 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

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

As the table shows, 0 is the identity.

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

Order of group:

Order of elements:
Because every element is its own inverse,

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

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


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

Generating sets:
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.


Order 2:

(all subgroups generated by a single element)

Order 4:

(all subgroups generated by two elements)

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}.