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Find the number of elements of order $1$, $2$, $5$ and $10$ in $D_{10}$.

I want to decompose the group into $\mathbb{Z}_2 \times \mathbb{Z}_{10}$ and then use my knowledge of cyclic groups, but I know this is incorrect because $\mathbb{Z}_2 \times \mathbb{Z}_{10}$ is Abelian and $D_{10}$ is not. I have seen some references to Sylow's theorem on the website, but we haven't learned this yet. How to proceed?

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  • $\begingroup$ $D_{10}$ is quite a small group. You can take each element in turn, and find its order, and that won't take you very long. $\endgroup$ Mar 18, 2018 at 18:02
  • $\begingroup$ anyway that's not brute force? $\endgroup$
    – yoshi
    Mar 18, 2018 at 18:08

1 Answer 1

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The answer to this was to recognize that all elements were rotations or order 2 flips. So count rotations as you would in $\mathbb{Z}_{2n}$, then add $n$ more order 2 elements (the reflections).

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