The Existence of a Subgroup Does every group of order $2k$, where $k$ is an odd positive integer, have a subgroup of order $k$?
 A: Consider the natural endomorphism $G \rightarrow S_{2k}$. By Cayley's Theorem $G$ has an element of order 2(the cycle decomposition of this element contains k transpositions thus is odd). Thus $ H = Im \, G \not\subset A_{2k}$. Then as $HA_{2k}=S_{2k}$ we have the following 
$$ \frac{\mid S_{2k}\mid }{\mid A_{2k} \mid } = \frac{\mid H \mid }{\mid H \cap A_{2k} \mid}$$
Thus $\mid H \mid / \mid H \cap A_{2k} \mid = 2$ and $H \cap A_{2k}$ is the desired subgroup.
A: Let G be a group of order 2k and then let G act on itself by the group operation. By the First Isomorphism Theorem, G is isomorphic to a subgroup of $S_{2k}$. Let that subgroup be H. Then define a homorphism $\theta:H\to C_2$ by sign. Then as G contains an element of order 2 by Cauchy's Theorem and it doesn't fix any elements when acting on G, H must contain an element of odd sign and so exactly half the elements in H have odd sign (subgroups of $S_n$ either have no elements of odd sign or exactly half of the elements have odd sign). Therefore the kernel of $\theta$ is a normal subgroup of H of index 2 and so G must contain a normal subgroup of order 2.
