2
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We discussed this in class and it was seemingly just mentioned in passing. Is there a proof available to this?

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6
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For any subgroup $H$, $G$ is a disjoint union of all the cosets $g_iH$ (where $g_i$ are representatives). Since $|G:H|=2$, there are only two cosets: $H$ and $gH$, i.e. 'the other one'. Therefore $G\setminus H=gH$, the other coset.

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1
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Hint: can you find a bijection, even in the infinite case? [take an element not in $H$ ...]

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0
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If $H$ is a subgroup of $G$ of index $2$ then one of its cosets is $H$. Since all cosets have equal size there can only be another coset which is?

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