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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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    $\begingroup$ 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. $\endgroup$ Aug 27 '12 at 18:12
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Because this is the definition of having the same cardinality.

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  • $\begingroup$ I understand that $H$ and $aH$ will have the same cardinality, but if there was another element $b \in G$ such that there was another coset $bH = {bh \mid h \in H}$, then what says that the cardinalities of $aH$ and $bH$ are the same? $\endgroup$
    – achacttn
    Aug 27 '12 at 18:23
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    $\begingroup$ Never mind, understood. Thank you $\endgroup$
    – achacttn
    Aug 27 '12 at 18:35
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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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    $\begingroup$ Thank you, your second point is something I just realized (embarrassingly slowly I might add) $\endgroup$
    – achacttn
    Aug 27 '12 at 18:57
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    $\begingroup$ @achacttn - happens to everyone! $\endgroup$
    – Belgi
    Aug 27 '12 at 19:01

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