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I am trying to find a non cyclic group of order 125 which has an element of order 25.

I know product of cyclic groups can be used, but I am not sure whether an element of order 25 exists.

is there any generalized method to come up with examples for any given group with any given order? like a non cyclic group of order 63, with element of order 21.

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  • $\begingroup$ Not sure if this would work, but maybe try a semi-direct product? $\endgroup$
    – Kenny Wong
    Commented Mar 6, 2017 at 9:04
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    $\begingroup$ What about $C_{25} \times C_5$ and $C_{21} \times C_3$? Consider the element $(1,0)$. $\endgroup$ Commented Mar 6, 2017 at 9:04
  • $\begingroup$ @SquirtleSquad Yeah you're right. $\endgroup$
    – Kenny Wong
    Commented Mar 6, 2017 at 9:07

1 Answer 1

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Group: $\mathbb{Z}_{25}\times\mathbb{Z}_{5}$

Order $25$ element: $(1,0)$

In general for non-cyclic group of order $pq$ with $p,q$ not coprime you can take $\mathbb{Z}_{p}\times\mathbb{Z}_{q}$ then $(1,0)$ will be of order $p$ as required.

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  • $\begingroup$ thanks. will there be element of order q in your general case? $\endgroup$
    – jnyan
    Commented Mar 6, 2017 at 9:13
  • $\begingroup$ yes, $(0,1)$, you could also just swap $p,q$ $\endgroup$ Commented Mar 6, 2017 at 9:19
  • $\begingroup$ thanks. it was really stupid of me. $\endgroup$
    – jnyan
    Commented Mar 6, 2017 at 9:20

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