Thread: Union of distinct subset in a finite group G.

If $G$ is a finite group, $G=A \cup B$ and $A \cap B =\emptyset$ then $G\neq AB$ .

Help me proving this exersies please !

Thank you!

By AB you mean x\in G s.t x=a.b where a\in A, b\in B and the operation is the one in G?

Yes AB={a.b|a in A and b in B} , A and B are distinct subsets of group G.

If G = AB, then every element of G should be written as x=ab for a\in G and b\in G. for a\neq b since A\intersect B = empty. Consider the identity e of G. You cannot represent e as ab.
Is our assumption $A \cap B =\emptyset$ correct? or is $A \cap B = identity$?

In exercises we have that $A \cap B =\emptyset$

If G = AB, we need to prove both inequalities. AB \in G and G \in AB

Obviously, AB is contained in G, since every x in AB is of the form ab which is also in G since G is a group.
Is G in AB. No. Since e \in G but e is not in AB by the assumption A \intersect B = empty. Hence G \neq AB

I hope this answers your question.

why you say that e is not in A.B ??...

I am so sorry. Please ignore my answer. There is a flaw in my argument.