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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. – Davide Giraudo Jan 29 '12 at 14:25
@Davide Giraudo: That's correct! – spohreis Jan 29 '12 at 14:41
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. – Jack Schmidt Jan 29 '12 at 18:50
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. – Arturo Magidin Jan 29 '12 at 22:02
There is a lower bound, namely $\frac{|H||K|}{|H\cap K|}$. From page 41, (Hungerford's book Algebra) $[H \vee K: H]\geq [K:H\cap K]$. – spohreis Jan 31 '12 at 0:27

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