Let $G$ be a finite group. Suppose that $A$ and $B$ are to subsets of $G$. If $|G|<|A|+|B|$ prove that $$G=AB.$$
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$\begingroup$ Do you mean that $A$ and $B$ are subgroups of $G$, or just arbitrary subsets? $\endgroup$– Math1000Jan 5, 2015 at 2:08
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$\begingroup$ They ae arbitary subsets of $G$ $\endgroup$– erfan soheilJan 5, 2015 at 2:10
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1$\begingroup$ ... and by $AB$ I imagine you mean $\{ab\;:a\in A,\;b\in B\}$. Am I right? $\endgroup$– JoeJan 5, 2015 at 2:11
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$\begingroup$ Yes exactly i mean that $\endgroup$– erfan soheilJan 5, 2015 at 2:17
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3 Answers
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Let $g \in G$. Note that $|gB^{-1}| = |B|$ so $|A| + |gB^{-1}| > |G|$; in particular, $A\cap gB^{-1} \neq \emptyset$. Let $a \in A\cap gB^{-1}$, then $a = gb^{-1}$ for some $b \in B$. Therefore $g = ab \in AB$ so $G = AB$.
answered Jan 5, 2015 at 2:35
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Let $A = \{ a_1, \ldots, a_n \}$, and let $g \in G$ be arbitrary. Let $x_k = a_k^{-1} g$. Then $x_1, \ldots, x_n$ are distinct elements of $G$, and since the complement of $B$ in $G$ has less than $n$ elements, there exists $k$ such that $x_k \in B$. Then $g = a_k x_k \in AB$, so $G \subseteq AB$, proving the claim.
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$\begingroup$ How you claim thatmthe complement of $B$ in $G$ has less than $n$ elements? $\endgroup$ Jan 5, 2015 at 2:35
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If you can find valid $A$ and $B$ such that it is not all of $G$ there must be at least one element missing, say it is $g$, multiply $A$ by $g^{-1}$. Now $\{e\}$ is missing. That is, if there is a counterexample to the theorem there is a counterexample in which $e$ does not not belong to $AB$
But if the sum of the cardinalities is more than $G$ there must be one element in $A$ such that the inverse is in $B$, so this is impossible.
