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A $p$-group is a group where the order of every group element is a power of the prime $p$. The presentation of a dihedral group $D_n$ of order $2 n$ is as follows.

$$D_n = \langle x, y \mid x^n = y^2 = (xy)^2 = e, y x=y^{-1}x\rangle$$

So, $D_n$ can be a $p$-group only when $n = 2^m$, when $m \in \mathbb{Z}$ ,because that is when $x^{2^m} = y^{2^1} = e$ for $p = 2$.

Am I getting it right?

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    $\begingroup$ I think so...and it is pretty obvious, isn't it? Since you're already determining that one of the primes that divide the group's order is $\;2\;$ , the only option left for it to be a $\;p\,-$ group is a $\;2\,-$ group, which would make $\;n\;$ a power of $\;2\;$ . $\endgroup$ – DonAntonio May 1 '16 at 23:35
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Since by definition $D_n$ has an element of order $2$, $D_n$ is a $p$-group iff it is a $2$-group iff $2n$ is a power of $2$ iff $n$ is a power of $2$.

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