1
$\begingroup$

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

$\endgroup$
3
  • 3
    $\begingroup$ Note that a group of order $85$ must be cyclic because of $5\nmid 17-1$ $\endgroup$ – Peter Jul 29 '18 at 15:10
  • 3
    $\begingroup$ In fact, $\mathbb Z_{85}$ is isomorphic to $\mathbb Z_{17}\times \mathbb Z_5$ $\endgroup$ – Peter Jul 29 '18 at 15:13
  • 1
    $\begingroup$ Oh youre right, thank you so much $\endgroup$ – J. González Jul 29 '18 at 15:25
1
$\begingroup$

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 $\Bbb Z_{85}.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.