an abstract algebra question: how to prove there exists a subgroup whose order is K If k is an odd and positive integer, please prove that any group whose order is 2k has a subgroup whose order is k.
I have been struggling for this, I tried to use mathematical induction to prove it, but maybe it doesn't work. Can someone help me solve this question? thank you for your help.
 A: Consider the left regular action of $G$ on $G$. When seen as a homomorphism of $G$ to $S_G$ the  image of $g$ is an odd permutation if and only if $\frac{|G|}{|g|}$ is odd. Notice when $G$ is of order $2k$ there is an element of order $2$ by Cauchy. So the image of the homomorphism has odd permutations. A subgroup of a symmetric group can half either exactly half odd permutations or no odd permutations.
Since in this case there is at least one odd permutation you can consider the preimage of the even permutations and use the first iso theorem to prove it has the desired order.
A: Let $G$ be a group of order $2k$ where $k$ is odd. By Cayley theorem we can see $G$ as a subgroup of $S_{2k}$.
Let $a\in G$ such that $|a|=2$ and $G=\{g_1,g_2,...,g_{2k}\}$ then notice that $a$ correspond to product of cylces $(g_1,ag_1)(g_2,ag_2),...(g_k,ag_k)$ thus $a$ correspond to  an odd permutation.
As $G $ is not contained in $A_{2k}$ then $GA_{2k}=S_{2k}\implies G/(A_{2k}\cap G)\cong S_{2k}/A_{2k}=Z_2$
Thus, $G\cap A_{2k}$ has index $2$ in $G$.
