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Given

  • $G$ be a finite group
  • $X$ is a subset of group $G$
  • $|X| > \frac{|G|}{2}$

I noticed that any element in $G$ can be expressed as the product of 2 elements in $X$. Is there a valid way to prove this?

If the third condition was $|X| = \frac{|G|}{2}$ instead, does the above statement still hold?

Thank you.

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Note that for any $g \in G$ the set $g X^{-1}$ has also more than $\frac{|G|}{2}$ elements, hence $gX^{-1} \cap X \neq \emptyset$ so that there are $x,y \in X$ such that $gx^{-1} = y$, i.e. $g = yx$.

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  • $\begingroup$ Thank you very much. I'm just wondering that, does the set $gX^{-1}$ have the same number of elements as the set $X$? $\endgroup$ – Alicia Hargrove Oct 22 '14 at 15:04
  • $\begingroup$ Yes, it does. the map $X \rightarrow gX^{-1}, x \mapsto gx^{-1}$ is a bijection with inverse $a \mapsto a^{-1}g$. $\endgroup$ – Matthias Klupsch Oct 23 '14 at 6:17
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Answer for the additional question: the statement need not be true if $|X|=|G|/2$.

For example, let $G$ be the cyclic group of order $2$ and $X$ consist of the non-identity element.

A (slightly) more general example: let $G$ be the cyclic group of order $2n$, let $g$ be a generator, and let $$X=\{g^{2k}\mid k=0,1,\ldots,n-1\}\ .$$

And a bit more general again: if $X$ is a proper subgroup (not just a subset) of $G$ then the product of elements of $X$ is still in $X$, and hence not all elements of $G$ will be obtained.

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  • $\begingroup$ For the first example with $|G| = 2$: taking $X$ to be the identity element works equally well (and perhaps even better, since the identity is idempotent) $\endgroup$ – zcn Oct 21 '14 at 7:37
  • $\begingroup$ Note that your second example can also be written as $X=\langle g^2\rangle$, i.e. the cyclic subgroup generated by $g^2$. $\endgroup$ – Mario Carneiro Oct 21 '14 at 17:58

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