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I know that if $H$ and $K$ are subgroups of a finite group $G$, then $|HK|=\frac{|H||K|}{|H\cap K|}$.

Is there a formula for $|\langle H\cup K\rangle|$, if $H$ and $K$ are nonempty subsets of a finite group $G$?

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    You are looking for a formula which involve the cardinal of $H$ and $K$ and potentially the cardinal of a set combination between $H$ and $K$, right? For example, if $H$ is a singleton, and we take $H=K$, $|\langle H\cup K\rangle|$ is the order of the element.2012-01-29
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    @Davide Giraudo: That's correct!2012-01-29
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    Consider the case when *H* and *K* are subgroups of order 2 with an intersection of size 1. The size of the subgroup generated by *H* and *K* is an arbitrary even number ≥ 4. I'm not sure what sort of formula you could be looking for under these circumstances.2012-01-29
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    Every countable group can be embedded in a $2$-generator group. In particular, $2$-generator groups can have arbitrarily large size. So if $H$ and $K$ are sets of size $1$, the size of $\langle H\cup K\rangle$ could potentially be as large as you want.2012-01-29
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    There is a lower bound, namely $\frac{|H||K|}{|H\cap K|}$. From page 41, (Hungerford's book Algebra) $[H \vee K: H]\geq [K:H\cap K]$.2012-01-31

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