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This seems like something that should be obviously true, yet when my friend asked me how I knew it, I couldn't come up with a proof. And now I'm wondering if it's possible that this could somehow turn out to be false.

Suppose I have groups $A$ and $A'$ with respective subgroups $B$ and $B'$, and bijections $\phi:A\to A'$ and $\varphi:B\to B'$.

Can I say that the sets $A/B$ and $A'/B'$ are in $1-1$ correspondence? I've found that what I thought would be the natural bijection $a + B\mapsto \phi(a) + B'$ need not even be well-defined. Unfortunately this kills a two page argument that took me a whole day to write. :( But all I need is a bijection, not an isomorphism. Can anyone help?

(fingers crossed)

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    Distinct left (or right) cosets of a subgroup are disjoint, so the coset space is a partition of the group. Each coset has the same number of elements.2012-09-18

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If $A$, $A'$ can be infinite, then this is false. Consider $A=A'=\mathbb{Z}$, $B=\mathbb{Z}$, $B'=2\mathbb{Z}$.

If $A$ and $A'$ are finite, then there is a bijection between $A/B$ and $A'/B'$, because both sets have size $\frac{\#A}{\#B}=\frac{\#A'}{\#B'}$

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    Yes this is what led me to jump to the incorrect conclusion.2012-09-19