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If $A$ and $B$ are finite subgroups, of orders $m$ and $n$, respectively, of the abelian group $G$, Prove that $AB$ is a subgroup of order $mn$ if $m$ and $n$ are relatively prime. Definition: $AB=\{ab|a\in A, b\in B \}$


My attemptation: $A$ has $m$ elements, $B$ has $n$ elements. There are $mn$ representations of elements in $AB$. If some of them are equal or there exist $a,a'\in A, b, b'\in B$ such that $ab=a'b'$, then $AB$ has elements less than $mn$, if such elements do not exist then $AB$ has order of $mn$. Therefore we want to show that if $m$ and $n$ are relatively prime then such elements do not exist. Please show me how to do it. I think it has something to do with Lagrange's theorem and properties of prime numbers.

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$AB $ is a subgroup because $G$ is abelian. We know that $$|AB| = \frac{|A| \cdot |B|}{|A \cap B|}$$ But if $\gcd(m,n) = 1$ then $|A \cap B| = 1 $ because $|A \cap B| \mid |A| = m$ and $|A \cap B| \mid |B| = n$ by Lagrange theorem. So $$|AB| = |A| \cdot |B| = mn$$

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  • $\begingroup$ >$|AB| = \frac{|A| \cdot |B|}{|A \cap B|}$ By what? $\endgroup$ – pxc3110 Apr 8 '14 at 6:50
  • $\begingroup$ @pxc3110: is a general formula very useful in group theory; probably you can find the proof on mathstack $\endgroup$ – WLOG Apr 8 '14 at 7:19
  • $\begingroup$ math.stackexchange.com/questions/168942/… $\endgroup$ – WLOG Apr 8 '14 at 7:21
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Hint: If $a_1b_1=a_2b_2$ then $a_2^{-1}a_1= b_2b^{-1}_1$. What can you say about the order of $a_2^{-1}a_1$ and of $b_2b^{-1}_1$?

[You also have to prove that $AB$ is indeed a subgroup.]

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