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Let $m$ be a positive integer. Prove that the dihedral group $D_{4m+2}$ of order $8m+4$ contains a subgroup of order 4.

Im fairly stumped as to where to start with this one, any input would be helpful. Thanks

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If you'd prefer not to use the Sylow theorems, we can also see this simply because $4m + 2 = 2(2m + 1)$. That is, if we present $D_n$ as $D_n = \left<\sigma,\tau\mid\sigma^n = \tau^2 = 1, \tau\sigma\tau = \sigma^{-1}\right>$, then if $n = 2k$ is even, we have $2k$ "rotation" elements ($\sigma^i$, $i = 0, \ldots, 2k - 1$) and $2k$ "reflection" elements ($\sigma^i\tau$, $i = 0, \ldots, 2k - 1$), so we can take $H = \{e, \sigma^k, \tau, \sigma^k\tau\}$ as the subgroup of order $4$, which is easily verified to be a subgroup.

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  • $\begingroup$ @Stahl I havent seen sylows theorem yet so this is of great help! $\endgroup$ – bobdylan Feb 21 '13 at 15:36
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Since it is of order $4(2m+1)$, it contains a Sylow-2-subgroup, of order =$4$.

Edit:
The first Sylow-theorem states that, if $p^m$ is the highest power of $p$ which divides the order of $G$, then $G$ contains a subgroup of order $p^m$. In our case, the highest power of $2$ dividing the order of $G$ is simply $4$, hence the result.

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  • $\begingroup$ Am I being ambiguous? $\endgroup$ – awllower Feb 21 '13 at 4:42
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    $\begingroup$ Your answer looks good to me. I've provided an alternative solution that works without Sylow theory, in case @bobdylan hasn't seen those theorems yet (although it assumes knowledge of a certain presentation of $D_n$). $\endgroup$ – Stahl Feb 21 '13 at 4:46

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