Does a group of order $400$ always have a subgroup of order $200$? 
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.
 A: Ancientmathematician has answered your first question. You have asked for a simple criterion to determine when a group of even order has a subgroup of index $2$. One such criterion is that the highest power of $2$ which divides $|G|$ is $2^1$, as mentioned in the comments.
Call $n \in \mathbb{N}$ a supersoluble number if every group of order $n$ is supersoluble. It is a standard theorem that a (finite) supersoluble group has subgroups of each possible order. In particular, if $G$ is a supersoluble group of even order then $G$ has a (necessarily normal) subgroup of index $2$. 
A theorem which can act as a supplement to the "$m=\text{odd}$" criterion is the following one. First, define $\psi$ to be 
the multiplicative function defined on prime powers as
$\psi(p^k) = (p^k-1)(p^{k-1}-1)\dots(p-1)$.
Theorem: Let $n=p_1^{a_1}p_2^{a_2} \dots p_r^{a_r}$ be a positive integer, where $p_1 < p_2 < \dots < p_r$.  Then $n$ is a supersoluble number if and only if:


*

*For all $1 \leq i \leq r$, the distinct prime factors of $\gcd(n,\psi(p_i^{a_i}))$ are the
same as those of $\gcd(n,p_i-1)$.

*If there exists $i \neq k$ such that $p_i \leq a_k$ (i.e. the value of some prime factor of $n$ is
less than the multiplicity of another), then
(a) There does not exist a prime $p_j$ such that $p_i | p_j-1$ and $p_j | p_k-1$, and
(b) $a_i \leq 2$, and if $a_i = 2$, then $p_i^2 | p_k - 1$.  
Applying this theorem to $n=20$ gives at once that $20$ is a supersoluble number, thus a fortiori every group of order $20$ has a subgroup of index $2$. Note here that the "$m=\text{odd}$" criterion tells you nothing about $n=20$. On the other hand, $n=12$ is not a supersoluble number because $\gcd(12,\psi(2^2)) =3$, whereas $\gcd(12,1) =1$. Indeed, $A_4$ is not supersoluble and it so happens that $A_4$ has no subgroup of index $2$.
Warning: if $n$ is not a supersoluble number, it does not follow that some group of order $n$ fails to have a subgroup of some order $d \mid n$ (much less specifically $d=\frac{n}{2}$). James' comment provides a concrete counterexample: every group of order $224=2^5 \cdot 7$ has subgroups of each possible order (i.e. $224$ is a Lagrangian number), but there are groups of order $224$ which are not supersoluble. I don't know if there are any useful arithmetic criteria guaranteeing that a number $n$ is Lagrangian.
Added: There is another criterion you can use which subsumes the "$m=\text{odd}$" one. Recall the definition of $\psi$ above. If $G$ is a group of order $n$ and if $\gcd(n,\psi(2^a))=1$, where $2^a$ is the highest $2$-power dividing $n$, then $G$ is $2$-nilpotent (see e.g. Isaacs' Finite Group Theory book, Corollary $5.29$). It is easy to see that a $2$-nilpotent group will always have a subgroup of index $2$. 
A: Let me also add a characterization for a group to have (or not to have) a subgroup of index $2$. One can show the following:
A group $G$ has no index $2$ subgroup iff $G$ is generated by squares, i.e. $G = \langle \lbrace g^2 \mid g \in G\rbrace \rangle $.
The group $\langle \lbrace g^2 \mid g \in G\rbrace \rangle \subset G$ is actually always a normal subgroup, but one can get some additional information if it has index $2$ in $G$:
A group $G$ has exactly one subgroup of index $2$ iff $\langle \lbrace g^2 \mid g \in G\rbrace \rangle $ has index $2$ in $G$.
A: 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
$$
\left\{ \begin{pmatrix} \alpha & 0 \\u & \alpha^{-1}  \end{pmatrix}\mid u\in GF(16), \alpha\in GF(16)^{*}, o(\alpha)|5 \right\}.
$$
