Specific question about the first omega subgroup of the non-abelian kernel of a frobenius group  Let $G=KH$ be a frobenius group with non-abelian kernel $K$,
$|H|=r-1$,
$|K|=r^2$,
$r=2^m$ for some odd integer $m$,
$Z(K)=K'=\Phi(K)$, the Frattini subgroup of $K$,
$[K:K']=|K'|=r$.
Let both $K/K'$ and $K'$ be elementary abelian 2-groups.
My questions are:
1) In my special case, is it correct that <{$k\in K| k^2=1$}$>=:\Omega_1(K)\neq K$ ?
2) I don't know much about $\Omega_1(K)$. Concerning my first question, are there any helpful theorems?
 A: Yes, this is true.  Suppose that $\Omega_1(K)=K$, and let $t\in K$ be such that $t^2=1$, but $t\notin Z(K)$.  Then $M=\langle t,Z(K)\rangle$ is an elementary abelian subgroup.  But $K=\cup_{h\in H}\ M^h$, because $H$ acts transitively on $K/\Phi(K)$, and thus every element of $K$ has order $2$, which is absurd.
Note that the above shows more; namely, in your case $\Omega_1(K)=Z(K)$.
It is key that you have this group $H$ of automorphisms.  For example, there is a group $L$ of order $64$ such that $Z(L)=\Phi(L)=L'$ is elementary abelian of order $8$, and $\Omega_1(L)=L$.
There are lots of things one can say about $\Omega_1(G)$ for p-groups $G$.  It actually doesn't seem that rare that $G=\Omega_1(G)$, though I believe it is slightly more restrictive in the $p=2$ case.  Such a group is called one-stepped, after Ito, and in the 2-group case this means there is a collection of involutions $t_1,\ldots,t_n$ - with $|G|=2^n$ - such that $\langle t_1,\ldots,t_i\rangle$ has order $2^i$.
A very good reference for all of this is the two-volume Groups of Prime Power Order by Berkovich and Janko.
