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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?

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1 Answer 1

up vote 2 down vote accepted

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.

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Thank you very much, it is clear to me now. Also thank you for mentioning theses books. –  Bernhard Boehmler Nov 2 '11 at 13:11
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