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.


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.