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I have to prove that a group of order $340$ has a cyclic normal subgroup of order $85$. Because $$340=17\times5\times2^2$$ and due to the third Sylow theorem that there is only one $17$-Sylow subgroup and one $5$-Sylow subgroup, so they are normal. So the group of order $85$ is also normal.

The problem is that, I know that, it has to be isomorphic to $\Bbb{Z}_{85}$ or $\Bbb{Z}_{17} \times \Bbb{Z}_5$ and I don't know how to prove that it's the first one. Any advice? Thanks

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    $\begingroup$ Note that a group of order $85$ must be cyclic because of $5\nmid 17-1$ $\endgroup$
    – Peter
    Jul 29, 2018 at 15:10
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    $\begingroup$ In fact, $\mathbb Z_{85}$ is isomorphic to $\mathbb Z_{17}\times \mathbb Z_5$ $\endgroup$
    – Peter
    Jul 29, 2018 at 15:13
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    $\begingroup$ Oh youre right, thank you so much $\endgroup$
    – J. González
    Jul 29, 2018 at 15:25

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Since $\Bbb Z_{17}$ has an element $h$ of order $17$ and $\Bbb Z_5$ has an element $k$ of order $5,$ then $\langle h,k\rangle$ is an element of $\Bbb Z_{17}\times\Bbb Z_5,$ whose order is $\operatorname{lcm}(17,5)=85.$ Since we readily have $\bigl\lvert\Bbb Z_{17}\times\Bbb Z_5\bigr\rvert=85,$ then this tells us that $\Bbb Z_{17}\times\Bbb Z_5$ is a cyclic group of order $85,$ meaning that it's isomorphic to $\Bbb Z_{85}.$

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