I have shown that any group of order 35 is cyclic. I don't know what to do next.
EDIT: I am not asking how to show that any group of order 35 is cyclic, just trying to show that that implies its unique
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1If you have shown that, you are done. There is only one cyclic group for each order! – Mathematician 42 Dec 16 '17 at 07:29
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(i) prove the Sylow subgroup $S$ of order $7$ is unique (Sylow's third theorem).
(ii) deduce $S$ is normal in $G$.
(iii) prove that $G$ is a semidirect product of $S$ with its Sylow $5$-subgroup $T$.
(iv) Show that the only action of $T$ on $S$ is trivial.
(v) prove this semidirect product is direct.
Angina Seng
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In a much more general sense you have the following result,
Result: If $G$ is a group of order $pq$ for some primes $p$, $q$ such that $p>q$ and $q\not| \;(p-1)$ then $G \cong\Bbb Z/pq\Bbb Z$.
So every group of order $pq$ satisfying the above conditions will be isomorphic to $\Bbb Z/pq\Bbb Z$ and hence upto isomorphism only one such group exists.
Naive
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