# Algebra of Complexes of a Group

Let us consider the set of all complexes of a group $$G$$, which is nothing but a power set of $$G$$. Let it be denoted by $$P\left( G \right)$$. Now, we define three binary compositions in $$P\left( G \right)$$. The two compositions, namely union and intersection of sets, are familiar ones. They are how we define the multiplication of complexes.

__Multiplication of Complexes__

Let $$H$$ and $$K$$ be two complexes of a group $$G$$ whose composition has been denoted multiplicatively, then the product of $$H$$ and $$K$$ denoted by $$HK$$ is defined as

\[HK = \left\{ {hk:h \in H,\,\,k \in K} \right\}\]

In other words $$HK$$ is the set of all possible products of elements of $$H$$ with those of $$K$$. It is evident that $$hk \in HK$$.

\[ \Rightarrow h \in H,\,\,k \in K\]

\[ \Rightarrow h,\,k \in G\]

\[ \Rightarrow hk \in G\]

\[HK \subset G\]

Thus the product of two complexes is also a complex of the group.

__Multiplication of Complexes is Associative__

Let $$H,K$$ and $$L$$ be three complexes of a group $$G$$ whose composition is denoted multiplicatively. Then

\[HK = \left\{ {hk:h \in H,\,\,k \in K} \right\}\]

\[\left( {HK} \right)L = \left\{ {\left( {hk} \right)l:h \in H,\,k \in K,\,l \in L} \right\} = \left\{ {h\left( {kl} \right):h \in H,\,k \in K,\,l \in L} \right\}\]

Because $$\left( {hk} \right)l = h\left( {kl} \right)$$, multiplication in $$G$$ is associative.

Also \[H\left( {KL} \right) = \left\{ {h\left( {kl} \right):h \in H,\,k \in K,\,l \in L} \right\}\]

\[\left( {HK} \right)L = H\left( {KL} \right)\]

__Inverse of Complexes in a Group__

Let $$H$$ be any complex of $$G$$ and let us define

\[{H^{ – 1}} = \left\{ {{h^{ – 1}}:h \in H} \right\}\]

Then $${H^{ – 1}}$$ is the complex of $$G$$ consisting of the inverse of the elements of $$H$$. This $${H^{ – 1}}$$ is called the inverse of complex $$H$$.