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There is a lemma that says that all left cosets $aH$ of a subgroup $H$ of a group $G$ have the same order.

The proof given is as follows...

The multiplication by $a \in G$ defines the map $H \rightarrow aH$ that sends $h\mapsto ah$. This map is bijective because its inverse is multiplication by $a^{-1}$.

I don't quite understand the proof. Why does having a bijective map mean that all sets of left cosets have the same order? Thank you

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    The order of a coset is defined as its cardinality. Two sets have the same cardinality if and only if there is a bijection between them.2012-08-27

2 Answers 2

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Because this is the definition of having the same cardinality.

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    Never mind, understood. Thank you2012-08-27
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First, you should know that two sets $A,B$ have the same size (by definition) if there is a bijection $f:A\to B$ (in this case there is also a bijection $g:B\to A$).

You can understand why this is the definition in the case that $A,B$ are finite by using a drawing.

Second, $|aH|=|H|=|bH|$ (since the size of every coset is equal to the size of $H$) and thus all cosets have the same cardinality.

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    @achacttn - happens to everyone!2012-08-27