Now we can use the n-chains to define our chain groups. We’re going to be using coefficients from $\mathbb Z_2$, which is a field, and remember there are only 2 elements ${0,1}$ where $1+1=0$.
The 0-chain group is defined as: Remember a group only has an addition operation defined but we’re building the group by using a multiplication operation from the field $\mathbb Z_2$. So this group is actually isomorphic to $\mathbb Z_{2}^{3} = Z_{2} \oplus Z_{2} \oplus Z_{2}$.
But we also want to represent our chain groups as a vector space. This means it becomes a structure where elements can be scaled up or down (i.e. multiplication operation) by elements of a field (in our case $\mathbb Z_2$) and added together, with all results still contained in the structure. If we only pay attention to the addition operation then we’re basically looking at its group structure, whereas if we pay attention to both the multiplcation and addition operations then we are considering it as a vector space.
0-chain vector space is generated by: (Yes it’s the same set that forms the group from above).
The vector space is the set of elements we can multiply by 0 or 1, and add them together. For example, we can do: $1(a,0,0) + 1(0,0,c) = (a,0,c)$. This vector space is so small $(2^3=8\ \text{elements})$ we can actually list out all the elements. Here they are:
You can see that we can add any two elements in this vector space and the result will be another element in the vector space. A random example: $(a,0,c) + (a,b,c) = (a+a,0+b,c+c) = (0,b,0)$. Recall addition is component-wise. We can also multiply vectors by an element in our field $\mathbb Z_2$ but since our field is finite with only 2 elements, it’s not very interesting, e.g. $1(a,b,0) = (a,b,0)$ and $0(a,b,0) = (0,0,0)$, but none the less, the multiplication operation still results in an element within our vector space.