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!

Attachment 25991

Printable View

- Nov 30th 2012, 05:39 AMjohn27Union 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!

Attachment 25991 - Nov 30th 2012, 07:00 AMcoolgeRe: Union of distinct subset in a finite group G.
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?

- Nov 30th 2012, 07:05 AMjohn27Re: Union of distinct subset in a finite group G.
Yes AB={a.b|a in A and b in B} , A and B are distinct subsets of group G.

- Nov 30th 2012, 07:10 AMcoolgeRe: Union of distinct subset in a finite 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 $A \cap B = identity$? - Nov 30th 2012, 07:21 AMjohn27Re: Union of distinct subset in a finite group G.
In exercises we have that $A \cap B =\emptyset$

- Nov 30th 2012, 07:54 AMcoolgeRe: Union of distinct subset in a finite group G.
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. - Nov 30th 2012, 09:01 AMjohn27Re: Union of distinct subset in a finite group G.
why you say that e is not in A.B ??...

- Nov 30th 2012, 10:54 AMcoolgeRe: Union of distinct subset in a finite group G.
I am so sorry. Please ignore my answer. There is a flaw in my argument.