Does a group of order $400$ always have a subgroup of order $200$?
I was considering some simple applications of Sylow theorems. I have made some questions. Among them, there was one which asks to prove that a group of order 200 always have a subgroup of order 100. This question has a simple solution since the Sylow group of order $25$ is unique and hence normal. After this, I became curious about the existence of normal subgroups of order $2^k p^l$ of groups of order $2^{k+1} p^l$.
The following two paragraphs are about my failed attempt which may not helpful.
For a group of order $400$, say $G$, let $n_5$ be the number of Sylow group of order $25$. If $n_5 =1$, we are done. So suppose that $n_5 = 16$. Let $A$ and $B$ be two Sylow subgroups of order $25$. Analysing the number of elements in the set $AB$, one can conclude that the order of $A \cap B$ to be $5$. This forced $A \cap B$ to be normal subgroup in $A$ and $B$. I was considered the normalizer $N(A \cap B)$. For this should contain $A$ and $B$, the order of $N(A \cap B)$ should be $200$ or $400$. If it was $200$, we are done. So suppose that for every pair of different Sylow groups of order $25$ $A$ and $B$, $A \cap B$ is normal. If there occur two different intersections of order $5$, we are done since the product of two such normal subgroups of order $5$ will provide a normal subgroup of order $25$.
The point I was stuck is there. Specifically, if a counterexample exists, it should have $16$ Sylow subgroups or order $25$ whose intersection is of order $5$.
The followings are just some consideration for generalization which may "not even wrong"
Most generally, I'm curious about the condition of $n$ which forces to exist a subgroup of order $n$ of a group of order $2n$.
Note that as all of you know, there are some groups of even order without having subgroups of index $2$.
All simple groups of even order have this property since a subgroup of index $2$ is automatically normal.
Anyway, let me just state the general question.
Find a simple criterion for a positive integer $n$ to have the property that every group of order $2n$ has a subgroup of index $2$.
Finally, thank you for your attention.
Best Answer
There is a group $H$ of order 80 whose normal subgroups have orders 1,16,80. Taking $G=\mathbb{Z}_5\times H$ gives you a group of order 400 with no normal subgroup of index 2.
To get an explicit version of $H$, just take the group of matrices $$ \left\{ \begin{pmatrix} \alpha & 0 \\u & \alpha^{-1} \end{pmatrix}\mid u\in GF(16), \alpha\in GF(16)^{*}, o(\alpha)|5 \right\}. $$