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

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

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.
Since it is of order $4(2m+1)$, it contains a Sylow-2-subgroup, of order =$4$.
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.
• 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 $D_n$). – Stahl Feb 21 '13 at 4:46