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Problem: Prove that if $G$ is a finite group, then the number of elements in the double coset $AxB$ is $\frac{|A| * |B|}{|A \cap xBx^{-1}|}$.

So here's what I know, and I don't know if these pieces of information will be useful. $|AB| = \frac{|A| * |B|}{|A \cap B|}$ and $[G:B] = [G:A][A:B]$.

I am especially caught up on the $xBx^{-1}$ part. I understand that $xBx^{-1} = B$ if the left and right cosets are equal (that's how the professor explained it anyway) and I know that the left and right cosets are equal if the subgroup is normal. I don't even know if that is useful for this problem though.

I really don't have any level of understanding when it comes to cosets beyond the definitions as my professor has really not taught this and the textbook does not seem very clear. There was a preceding problem and the link will be provided in case that is necessary to prove this problem.

Equivalence class of double coset

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    **Hint:** $|AxB|=\big|A\left(xBx^{-1}\right)\big|$. – 2017-01-29
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    Hint #2: even if $B$ is *not* normal, $xBx^{-1}$ is still a subgroup, call it $C$ (you should prove this to your own satisfaction). Show that $b \mapsto xbx^{-1}$ is a bijection $B \to C$. You already know what $|AC|$ is. – 2017-01-29
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    So would that then lead to something like this? $|A(xBx^{-1})|=\frac{|A|*|xBx^{-1}|}{|A\cap xBx^{-1}|}$. Or am I doing something wrong here? – 2017-01-29
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    Yes, that's the right idea, and you just need to show that $|B| = |xBx^{-1}|$, which shouldn't be too hard (use the same idea you used to show all left (or right) cosets have the same size). – 2017-01-29
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    Oh, so it's not showing that $B = xBx^{-1}$, it's showing that $|B| = |xBx^{-1}|$. Thanks for the assist. – 2017-01-29

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Decompose $AxB = \amalg_{1\leq i \leq n} Axb_i$ , and $|AxB|=|A|\times n$.

Noticing $B= \amalg_{1\leq i \leq n} (B \cap x^{-1}Ax)b_i$ , thus $n = \frac{|B|}{|B \cap x^{-1}Ax|}$, $|AxB|=\frac{|A|\times |B|}{|B \cap x^{-1}Ax|}$.

Order of Double Coset