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**Showcase_22** Let $\displaystyle |G:H|=|G|/|H|=n$ for $\displaystyle n \in \mathbb{N}$.

Let $\displaystyle H=\{h^0, \ldots , h^{n-1} \}$ for some $\displaystyle h \in H$.

Uh? You can't assume this unless you know $\displaystyle H$ is cyclic (which isn't a given) AND $\displaystyle h$ is one of its generators...

Then $\displaystyle Hg=\{h^0g, \ldots , h^{n-1}g \}$ for $\displaystyle g \in G$.

$\displaystyle (Hg)^{-1}=\{(h^0g)^{-1}, \ldots , (h^{n-1}g)^{-1} \}$

This is equal to $\displaystyle (Hg)^{-1}=\{g^{-1}(h^0)^{-1}, \ldots ,g^{-1}(h^{n-1})^{-1} \}$________________(*)

Since inverses are unique, and $\displaystyle H$ is a group (closure axiom) we have thatb $\displaystyle (h^k)^{-1}=h^i$ for $\displaystyle 0 \leq k \leq n-1$ for some $\displaystyle 0 \leq i \leq n-1$.

Hence we have $\displaystyle (Hg)^{-1}=\{g^{-1}h^0, \ldots , g^{-1}h^{n-1} \} =g^{-1}H$

Since $\displaystyle |G:H|=n$, we have exactly $\displaystyle n \ g^{-1}$'s to multiply $\displaystyle H$ by.

Thus, we have $\displaystyle n$ left cosets of $\displaystyle H$ in $\displaystyle G$.

I think this is right (can someone check =S), but step (*) applies only to multiplicative groups. Is it a fact that addition is always commutative?