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I'm confused about a solution to the question, "Does a group of order $35$ contain an element of order $5$? of order $7$?"

Aren't the cosets $$H, gH, g^2H,\dots, g^6H$$ disjoint by definition of a coset, in that they partition $G$? Why is the extra step necessary?

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  • $\begingroup$ I'm not sure why you edited out the image. It removes the context of the question, and your question seems unrelated to the first line, $\endgroup$ – Matt Samuel Feb 16 '16 at 1:35
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Cosets corresponding to different elements need not be disjoint (and if they are not disjoint, then they are the same). Indeed, multiplying on the right by any element of $H$ preserves the coset. Further, if $g$ had order $5$, then we'd have $gH=g^6H$ because $g^6=g$.

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It can turn out to be the case that $g \notin H$, but $g^n \in H$ for some integer $n$. Take the quaternion group for example, with generators $i$ and $j$ of order $4$. Then $i \neq j$, but $i^2 = j^2$.

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