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?



Specific question about the first omega subgroup of the non-abelian kernel of a frobenius group 
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.


