4
$\begingroup$

This question already has an answer here:

I wish to prove the exercise which states that for a set $A$ containing more than half elements of a group $G$, every element of $G$ is a product of two elements of $A$.

My attempt:

By Lagrange Theorem, subgroup generated by $A$ must coincide with $G$, so every element in $G$ is a product of elements from $A$. How can I prove that the product is of exactly two elements?

$\endgroup$

marked as duplicate by Derek Holt group-theory Feb 7 '16 at 18:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

4
$\begingroup$

Let $g\in G$. Then $A$ and $gA^{-1}$ are two sets larger than half of $G$, hence intersect. But if $a_1=ga_2^{-1}$ then $g=a_1a_2$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.