# 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 $H$.