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According to my lecture notes, for a subgroup $H$ of a group $G$, the (right) cosets of $H$ in $G$ are all the sets given by $$ Hx = \{hx: h \in H\} $$ Where $x \in G$.

This implies that the number of cosets would be given by $[G : H]= |G|$.

However, Lagrange's Theorem states that $[G:H]=\frac{|G|}{|H|}$.

Why is this the case?

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When you list the right cosets as $Hx$ for all $x \in G$ there are repetitions: we have an equality $Hx = Hy$ whenever $xy^{-1} \in H$. In particular, there are precisely $|H|$ repetitions of the identity coset $H = He$ that appear in the list. This is a general phenomenon, as evidenced by the fact that all the cosets are in bijective correspondence, and so you have to divide $|G|$ by $|H|$ to account for the repetitions.

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  • $\begingroup$ This is essentially the answer I would write, so I'll just add that we can phrase this as saying that the function \begin{align*} f: G &\to G/H \\ x &\mapsto xH \end{align*} is $\lvert H \rvert$-to-one, rather than one-to-one. $\endgroup$
    – pjs36
    Dec 20 '15 at 2:38
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Technically, it makes no sense fora set of cosets to be of cardinality even bigger then $G$, since this set is intuitively equinumerous to a set consisting of different elements $K$, such that if $g_1,g_2\in K$, $g_1H\ne g_2H$.

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Have you noticed that every coset of a subgroup H (in your notation) of a group G has the same number of elements as your H subgroup? So, yes, for a finite group the "index" (G:H) = order of G / order of H. The answer to your question is: because if you look at every single coset at H they will all have |H| elements

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