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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. –  anon Sep 18 '12 at 1:36

1 Answer 1

up vote 8 down vote accepted

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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That does it. Darn. I'll have to come up with something else. Thanks! –  Kyle Schlitt Sep 18 '12 at 2:00
@lovinglifein2012: It should be pointed out that what you are saying does happen in the finite case. –  user1729 Sep 18 '12 at 12:14
Yes this is what led me to jump to the incorrect conclusion. –  Kyle Schlitt Sep 19 '12 at 23:42

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