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If $G$ is Abelian, and it has subgroups $|A|$ and $|B|$ aren't relatively prime (its means they have mutual divisor). what is $|AB|$?

NOTE: I succeeded in proving that if $|A|$ and $|B|$ were relatively prime (which means that $gcd (|A|,|B|)=1)$ then $|AB|=|A||B|$. now im stuck at this one, any kind of help would be appreciated.

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It depends on whether $A$ and $B$ have any elements in common (it's not enough to know just the numbers $|A|$ and $|B|$). The general result is $$ |AB| = \frac{|A|\cdot |B|}{|A\cap B|} $$ This is true even for non-abelian $G$, although in that case $AB$ need not be a subgroup, but rather just the set $\{ab\mid a\in A, b \in B\}$.

Note that $A\cap B$ is a subgroup of both $A$ and $B$, so if $|A|$ and $|B|$ are coprime, then by Lagrange's theorem we must have $|A\cap B| = 1$ and we recover your result.

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  • $\begingroup$ what is the name of this formula? and how can I find its proof ? $\endgroup$ – F1sargyan Apr 15 '16 at 15:24
  • $\begingroup$ @F1sargyan Google "product of subgroups" and you should get a few results. I don't think it has a name as such, though. A proof can be found here. $\endgroup$ – Arthur Apr 15 '16 at 15:27
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Since $G$ is abelian, the map $\mu\colon A\times B\to AB$ defined by $(a,b)\mapsto ab$ is a (surjective) group homomorphism, so $$ A\times B/\ker\mu \cong AB $$ So all you need to compute is $|\ker\mu|$. Now $(a,b)\in\ker\mu$ if and only if $b=a^{-1}\in A\cap B$. So…

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We have for every finite group $G$, and subgroups $A$ and $B$ the following formula:

$$ |AB|=\frac{|A|\cdot |B|}{A\cap B}. $$ Here $AB$ is a subgroup of $G$ if and only if $AB=BA$. If $|A|$ and $|B|$ are coprime then $|A\cap B|=1$ by Lagrange. If not, then $|AB|$ is just a divisor of $|A|\cdot |B|$, but need not be equal.

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  • $\begingroup$ But I said that they aren't coprime! $\endgroup$ – F1sargyan Apr 15 '16 at 15:22
  • $\begingroup$ what is the name of this formula? and how can I find its proof ? $\endgroup$ – F1sargyan Apr 15 '16 at 15:24
  • $\begingroup$ I did not say that they must be coprime. The formula is proved in many books on group theory. $\endgroup$ – Dietrich Burde Apr 15 '16 at 15:27

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