\documentclass [12pt] {article} \usepackage{latexsym} \newtheorem{thm}{Theorem}[section] \newtheorem{lem}[thm]{Lemma} \newtheorem{cor}[thm]{Corollary} \newtheorem{prop}[thm]{Proposition} \newtheorem{defn}{Definition}[section] \newtheorem{ex}{Example}[section] \newtheorem{exs}{Examples}[section] \newtheorem{rem}{Remark}[section] \newtheorem{rems}{Remarks}[section] \newcommand{\Prf}{\noindent{\it Proof.}\quad } \newcommand{\qed}{\hfill$\Box$} \setlength{\textheight}{9.0in} \setlength{\textwidth}{6.1in} \addtolength{\topmargin}{-0.5in} \addtolength{\oddsidemargin}{-0.5in} \newfont{\sss}{cmss12 scaled 1000} \title{Finite 2-Groups $G$ with $|\Omega _3(G)|\leq 2^5$} \author{ Zdravka Bo\v{z}ikov \\ Faculty of Civil Engineering, University of Split, \\ 21000 Split, Croatia \\ {\tt E-mail: Zdravka.Bozikov@gradst.hr} \\ Zvonimir Janko \\ Mathematical Institute, University of Heidelberg, \\ 69120 Heidelberg, Germany \\ } \date{} \begin{document} \maketitle \begin{abstract} In this paper all finite 2-groups $G$ are determined up to isomorphism with the property $G>\Omega _3(G)$ and $|\Omega _3(G)|\leq 2^5$. Also, we show that any finite 2-group $G$ containing exactly one subgroup of order $2^{n+2}$ and exponent $\leq 2^n$ satisfies $|\Omega _n(G)|= 2^{n+2}$, where $n\geq 2$ is any fixed integer. \end{abstract} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Introduction and preliminary results} $\,\,\,\,\,\,\,$ The ultimate goal is to classify finite 2-groups $G$ of order $>2^6$ with $|\Omega _3(G)|=2^6$ (Berkovich [1, Research problem 526]). However, if $\Omega _3(G)\not\leq \Phi (G) $, there is a maximal subgroup $M$ of $G$ such that $|\Omega _3(M)|\leq 2^5$ and without the exact knowledge of the structure of $M$ there is no chance to determine $G$ in that case. In this paper we shall determine up to isomorphism all 2-groups $G$ with $|\Omega _3(G)|\leq 2^5$ and $G>\Omega _3(G)$. The cases $|\Omega _3(G)|=2^3$ and $|\Omega _3(G)|=2^4$ are easy (Proposition 1.7). If $|\Omega _3(G)|=2^5$, we show first that $exp(\Omega _3(G))=2^3$ and $|\Omega _2(G)|=2^4$ (Proposition 1.8). This allows us to use the classification of 2-groups $G$ with $|\Omega _2(G)|=2^4$ and $|G|>2^4$ given in Janko [4]. Indeed, in section 2 we consider non-metacyclic 2-groups $G$ and get our result almost immediately (Theorem 2.1). In section 3 we consider metacyclic 2-groups $G$ with the above property. This case is difficult since in Janko [4] there is no single information for metacyclic groups but we have to determine the groups $G$ up to isomorphism also in this case (Theorem 3.1). The corresponding results for $p>2$ seem to be much more difficult. Finally, in section 4 we classify the 2-groups $G$ containing exactly one subgroup of order $2^5$ and exponent $\leq 8$ (Berkovich [1, Research problem 437]). We show that such groups satisfy $|\Omega _3(G)|=2^5$. In fact, we get a more general result (Theorem 4.2). The groups considered are finite and our notation is standard. For the convenience we give here some known facts and then we prove two preliminary results (Propositions 1.7 and 1.8). \begin{prop} (Suzuki [7, \S 2, Exercise 2(c)]) Let $G$ be a nonabelian p-group. If $G$ has a subgroup $A$ of order $p^2$ such that $C_G(A)=A$, the $G$ is of maximal class. \end{prop} \begin{prop} (Berkovich [2, Lemma 2.1]) Let $G$ be a group of order $2^m,\,m\geq 4$, satisfying $|\Omega _2(G)|\leq 8$. Then one of the following holds: \begin{itemize} \item[(a)] $G\cong M_{2^m}=\langle a,b\,|\, a^{2^{m-1}}=b^2=1,\,\,a^b=a^{1+2^{m-2}}\rangle .$ \item[(b)] $G$ is abelian of type $(2^{m-1},2)$. \item[(c)] $G$ is cyclic of order $2^m$. \item[(d)] $G=\langle a,b\,|\, a^{2^{m-2}}=b^8=1,\,\,a^b=a^{-1},\,\,a^{2^{m-3}}=b^4,\,\,m\geq 5 \rangle .$ \end{itemize} \end{prop} \begin{prop} (Janko [4]) Let $G$ be a 2-group of order $>2^4$ and $|\Omega _2(G)|=2^4$. Then $\Omega _2(G)$ is isomorphic to one of the following groups: $$Q_8*C_4,\,\,Q_8\times C_2,\,\,D_8\times C_2,\,\,C_4 \times C_2\times C_2,\,\,and \,\,C_4\times C_4 . $$ If $\Omega _2(G)\cong Q_8\times C_2 $ or $\Omega _2(G)\cong D_8\times C_2 $, then $|G|=2^5$. \end{prop} \begin{prop} (Berkovich [2, Theorem 1.1, Remark 2]) If $G$ is a 2-group such that $\Omega _2(G)$ is metacyclic, then $G$ is also metacyclic. \end{prop} \begin{prop} Let $G$ be a p-group such that $|G'|=p$ and $d(G)=2$. Then $G$ is minimal nonabelian. \end{prop} \Prf Since $G$ is of class $2$, $\mho _1(G)\leq Z(G)$. Indeed, we have $[x^2,y]=[x,y]^2=1$ for any $x,y\in G$. It follows that $\Phi (G)=G'\mho _1(G)\leq Z(G)$ so $Z(G)=\Phi (G)$ has index $p^2$ in $G$. Hence each maximal subgroup of $G$ is abelian and we are done. \qed \begin{prop} (Miller-Moreno[5] and Redei [6]) Let $G$ be a minimal nonabelian p-group. If $G$ is metacyclic, then $$ G=\langle a,b\,|\,a^{p^m}=b^{p^n}=1,\,\,a^b=a^{1+p^{m-1}},\,\,m\geq 2,\,\,n\geq 1 \rangle $$ or $G\cong Q_8$ (the quaternion group). \end{prop} \begin{prop} Let $G$ be a 2-group with $G>\Omega _3(G)$. If $|\Omega _3(G)|\leq 2^3$, then $|\Omega _3(G)|= 2^3$ and $G$ is cyclic. If $|\Omega _3(G)|= 2^4$, then $\Omega _3(G)$ is abelian of type $(8,2)$ and $G$ is either abelian of type $(2^m,2)$ or $G\cong M_{2^{m+1}},\,\,m\geq 4$. \end{prop} \Prf Set $\Omega _3(G)=H$. If $exp(H)\leq 4$, then taking an element $x\in G-H$ with $x^2\in H$ we get $o(x)\leq 8$, which is a contradiction. Hence $exp(H)\geq 8$ and so $|H|\geq 2^3$. If $|H|=2^3$, then $|\Omega _2(G)|=4$ and Proposition 1.2 implies that $G$ is cyclic. Suppose $|H|=2^4$. Since $H$ is generated by its elements of orders $\leq 8$, $H$ is not cyclic but $H$ has a cyclic subgroup of index $2$. Suppose that $H$ is of maximal class. Then $H$ possesses a subgroup $K$ of order $4$ such that $C_H(K)=K$. If $C_G(K)>K$, then taking an element $y\in C_G(K)-K$ with $y^2\in K$ we get $o(y)\leq 8$, a contradiction. Thus $C_G(K)=K $ and Proposition 1.1 implies that $G$ is of maximal class. But then $\Omega _2(G)=G$ and since $\Omega _3(G)\geq \Omega _2(G)$, we have a contradiction. Hence $H$ is not of maximal class. It follows that $H$ is either abelian of type $(8,2)$ or $H\cong M_{2^4}$. In any case $\Omega _2(G)$ is abelian of type $(4,2)$ and we may apply again Proposition 1.2. In case (d) of that proposition, $\Omega _3(G)=G$ which contradicts our assumption. Hence $G$ must be isomorphic to a group (a) or (b) of that proposition. In both cases $H$ is abelian of type $(8,2)$ and we are done. \qed \begin{prop} Let $G$ be a 2-group with $G>\Omega _3(G)$. If $|\Omega _3(G)|= 2^5$, then $exp(\Omega _3(G))$ $=2^3$ and $|\Omega _2(G)|= 2^4$. \end{prop} \Prf Set $\Omega _3(G)=H$ so that $|H|=2^5$ and $G>H$. If $H$ is cyclic, then $\Omega _3(H)A$, then take an element $x\in C_G(A)-A$ such that $x^2\in A$. Then $o(x)\leq 8$, a contradiction. Hence $C_G(A)=A $ and Proposition 1.1 implies that $G$ is of maximal class. But then $\Omega _3(G)=G$, a contradiction. It follows that $H$ is either abelian of type $(16,2)$ or $H\cong M_{32}$. In any case, $\Omega _3(H)2^3$. Proposition 1.2 implies that $H/R$ is abelian of type $(4,2)$. In particular, $H'\leq R$ and $H'$ is elementary abelian. If the class $cl(H)$ of $H$ is $\leq 2$, then for each $x,y\in H$ with $o(x)\leq 4$, $o(y)\leq 4$, we get $$(xy)^4=x^4y^4[y,x]^6=1, $$ which implies that $exp(H)\leq 4$, a contradiction. Thus (since $H$ is not of maximal class) $cl(H)=3$ and so $H'=R$ and $R\not\leq Z(H)$. We set $C=C_H(R)$ and $D=C_G(R)$. Since $|H:C|=2$, $D$ covers $G/H$ and $G=HD$. If $exp(C)\leq 4$, then taking $x\in D-C$ such that $x^2\in C$, we get $o(x)\leq 8$, a contradiction. It follows $exp(C)=8$. Let $a\in C$ with $o(a)=8$. Then $C=R \langle a \rangle $, $|R\cap \langle a \rangle|=2$, and so $C$ is abelian of type $(8,2)$. Since $\Omega _3(D)=C $, our Proposition 1.7 implies that $D$ is either abelian of type $(2^m,2)$ or $D\cong M_{2^{m+1}}$, $m\geq 4$. In any case, $\Phi (D)$ is cyclic of order $2^{m-1}$ and so $\Phi (D)$ contains a characteristic cyclic subgroup $Z$ of order $8$. It follows that $Z\leq C$ and $Z$ is normal in $G$. But $|H/Z|=4$ and so $H'\leq Z$. This contradicts the fact that $H'=R$ and $|R\cap Z|=2$. We have proved that $\Omega _2(G)2^5$ with $|\Omega _3(G)|=2^5$. We set $H=\Omega _3(G) $ and apply Proposition 1.8. We see that $exp(H)=2^3$ and $\Omega _2(G)=K $ is of order $2^4$ so that $|H:K|=2$. By Proposition 1.4, $K$ is non-metacyclic. We are in a position to use Proposition 1.3. Since $|G|>2^5$, we get either $K\cong Q_8*C_4$ or $K\cong C_4\times C_2\times C_2$. All elements in $H-K$ are of order $8$. Assume first $K=Q*Z$, where $Q\cong Q_8$, $Z\cong C_4$, and $Q\cap Z=Z(Q)$. All elements in $K-(Q\cup Z)$ are involutions and so $Q$ is the unique quaternion subgroup in $K$. Thus $Q$ is normal in $G$ and so also $C=C_G(Q)$ is normal in $G$. Since $|G/(QC)|\leq 2$ and $|G|>2^5$, we have $|C|\geq 2^3$ and $C\cap K=Z$. Also, $\Omega _2(C)=Z $ and so Proposition 1.2 implies that $C$ is cyclic and $H=Q*\Omega _3(C) $. Suppose that $|G/(QC)|= 2$. Since $G/C\cong D_8$, there is an element $c\in G-(QC)$ such that $c^2\in C$. Set $P=C\langle c \rangle$. Again, $\Omega _2(P)=Z$ and so $P=\langle c \rangle $ is cyclic. But $\langle c \rangle$ induces on $Q$ an "outer" involutory automorphism and so we may set $Q=\langle a,b \rangle $ so that $a^c=a^{-1}$ and $b^c=ba$. Assume now $K\cong C_4\times C_2\times C_2$ so that $E=\Omega _1(K)$ is a normal elementary abelian subgroup of order $8$ in $G$. We have $\mho _1(K)=\langle z \rangle $ is a central subgroup of order $2$ in $G$. Obviously, $K/E$ is the unique subgroup of order $2$ in $G/E$ and so $G/E$ is either cyclic (of order $\geq 8$) or generalized quaternion. Suppose that $G/E$ is generalized quaternion and let $L/E$ be a cyclic subgroup of index $2$ in $G/E$. For each $x\in G-L$, $x^2\in K-E$ and so $o(x)=8$ which forces $\Omega _3(G)=G$, a contradiction. We have proved that $G/E$ is cyclic. Let $a\in G-K$ be such that $\langle a \rangle$ covers $G/E$. Then $\langle a \rangle$ is cyclic of order $2^m,\,m\geq 4$, $G=E\langle a \rangle $, $K\cap \langle a \rangle\cong C_4$, and $E\cap \langle a \rangle = \langle z \rangle $. We may set $E=\langle e,u,z \rangle $ so that we have the following possibilities for the action of $\langle a \rangle$ on $E$: \begin{itemize} \item[(i)] $e^a=eu$, $u^a=uz$ (here $a$ induces an automorphism of order $4$ on $E$); \item[(ii)] $e^a=e$, $u^a=uz$ (here $G=\langle e \rangle \times \langle a,u \rangle \cong C_2\times M_{2^{m+1}}$; \item[(iii)] $e^a=eu$, $u^a=u$ (here $G$ is minimal nonabelian); \item[(iv)] $[E,a]=1$ (here $G$ is abelian of type $(2^m,2,2)$). \end{itemize} We have proved the following result. \begin{thm} Let $G$ be a non-metacyclic 2-group of order $>2^5$ with $|\Omega _3(G)|=2^5$. Then one of the following holds. \begin{itemize} \item[(a)] $G=QP$, where $Q=\langle a,b \rangle $ is a normal quaternion subgroup of $G$, $P=\langle c\rangle $ is cyclic of order $2^m,\,m\geq 4$, $Q\cap P=Z(Q)$ and $c$ either centralizes $Q$ or $a^c=a^{-1}$ and $b^c=ba$. \item[(b)] $G=EP$, where $E=\langle e,u,z \rangle $ is a normal elementary abelian subgroup of order $8$, $P=\langle a\rangle $ is cyclic of order $2^m,\,m\geq 4$, and $E\cap P=\langle z \rangle $ with $z=a^{2^{m-1}}$. For the action of $\langle a\rangle$ on $E$ we have one of the following possibilities: \begin{itemize} \item[(i)] $e^a=eu$, $u^a=uz$, and here $a$ induces an automorphism of order $4$ on $E$; \item[(ii)] $e^a=e$, $u^a=uz$, and here $G \cong C_2\times M_{2^{m+1}}$; \item[(iii)] $e^a=eu$, $u^a=u$, and here $G$ is minimal nonabelian; \item[(iv)] $e^a=e$, $u^a=u$, and here $G$ is abelian of type $(2^m,2,2)$. \end{itemize} \end{itemize} \end{thm} \section{Metacyclic case} $\,\,\,\,\,\,\,$ Throughout this section we assume that $G$ is a metacyclic 2-group of order $>2^5$ with $|\Omega _3(G)|=2^5$. We set $H=\Omega _3(G) $. By Proposition 1.8, $exp(H)=2^3$ and $\Omega _2(G)=K $ is of order $2^4$. Proposition 1.3 implies that $K\cong C_4\times C_4$ and all elements in $H-K$ are of order $8$. It follows that $R=\Omega _1(K)$ is a normal 4-subgroup. Suppose that $C_H(R)=K$. Since $C_G(R)$ covers $G/H$, there is an element $x\in C_G(R)-H$ with $x^2\in K$. But then $o(x)\leq 8$, a contradiction. We have proved that $R\leq Z(H)$. Let $Y/K$ be a subgroup of order $2$ in $G/K$. Since $Y\leq \Omega _3(G)$, we get $Y=H$. Hence $H/K$ is the unique subgroup of order $2$ in $G/K$. This implies that $G/K$ is either cyclic (of order $\geq 4$) or generalized quaternion. We study first the case, where $G/K$ is cyclic. Let $a\in G-K$ be such that $\langle a\rangle$ covers $G/K$. Then $ \langle a\rangle\cap H\cong C_{2^3}$, $\langle a\rangle\cap K\cong C_{2^2}$, and $\langle a\rangle\cap R =\langle z\rangle\leq Z(G) $. Thus $a$ is of order $2^m,\,m\geq 4$, and we set $s=a^{2^{m-3}}$ and $v=a^{2^{m-2}}$ so that $s^2=v$, $v^2=z$, and $o(s)=8$. Set $C=C_G(K)$. Since $G/C$ is cyclic and acts faithfully on $K$, we have $|G/C|\leq 4$. For the structure of $Aut(C_4\times C_4)$ see Proposition 1.11 in Janko [3]. Also, $G'\leq K$ and $G'$ is cyclic which implies that $|G'|\leq 4$. If $G'=\{ 1\}$, then $G$ is abelian of type $(2^m,4),\,m\geq 4$. Suppose $|G'|=2$. Then Proposition 1.5 shows that $G$ is minimal nonabelian. Since $|G|>8$, Proposition 1.6 together with the fact that $|\Omega _3(G)|=2^5$ gives $$G=\langle a,b\,|\, a^{2^m}=b^4=1,\,\,m\geq 4 ,\,\, a^b=a^{1+2^{m-1}}\,\,or \,\,b^a=b^{-1} \rangle . $$ Suppose $|G'|=4$ and $|G:C|=2$. Then $a$ induces an involutory automorphism on $K$, $H=\langle K,s\rangle \leq C$, and so $H$ is abelian of type $(8,4)$. If $a$ centralizes $K/R$, then $G'\leq R$. But $G'$ is cyclic and so $|G'|\leq 2$, a contradiction. Hence $a$ acts non-trivially on $K/R$ and so $a$ acts also non-trivially on $R=\mho _1(K)$. There is an element $w\in K-R$ such that $w^2=u\in R-\langle z\rangle $, $w^a=w'$, $(w')^2=uz$, $K=\langle w\rangle\times \langle w'\rangle $, and $u^a=uz$. Since $a$ induces an involutory automorphism on $K$, $(w')^a=w^{a^2}=w$ and therefore $C_K(a)= \langle ww'\rangle = \langle v\rangle $ with $(ww')^2=z$. From $$w^aw^{-1}=[a,w^{-1}]=w'w^{-1}=w'wu=ww'u $$ follows that $G'= \langle ww'u\rangle \cong C_4 $, $(ww'u)^2=z$, and $G'\not\leq Z(G)$. From the above, $ww'=vz^{\epsilon}\,\,(\epsilon =0,1) $ and set $l=swu^{\epsilon +1}$. Then $$l^2=s^2w^2=vu=ww'uz^{\epsilon}\in G',\,\,(vu)^2=z, $$ and so $\langle l\rangle \cong C_8$ is normal in $G$ with $\langle l\rangle\cap \langle a\rangle = \langle z\rangle$. We compute $$l^a=(swu^{\epsilon +1})^a=sw'u^{\epsilon +1}z^{\epsilon +1}=s^{-1} s^2w'u^{\epsilon +1}z^{\epsilon +1} = $$ $$s^{-1}vw'u^{\epsilon +1}z^{\epsilon +1}= s^{-1}(ww'z^{\epsilon})w' u^{\epsilon +1}z^{\epsilon +1} = $$ $$s^{-1}w^{-1}w^2(w')^2u^{\epsilon +1}z=s^{-1}w^{-1}uuzu^{\epsilon +1}z = s^{-1}w^{-1}u^{\epsilon +1}=l^{-1} .$$ We have obtained the following metacyclic group $$G=\langle a,l\,|\,a^{2^m}=l^8=1,\,\,a^{2^{m-1}}=l^4, \,\,l^a=l^{-1},\,\,m\geq 4 \rangle , $$ where $H=\langle l,s\rangle =\langle l, a^{2^{m-3}}\rangle $ is abelian of type $(8,4)$. It remains to study here the case $|G'|=4$ and $|G:C|=4$. Hence $a$ induces on $K$ an automorphism of order $4$ and so, in particular, $a$ acts non trivially on $R$. This implies that $C_K(a)= \langle v\rangle $, where $\langle v^2\rangle = \langle z\rangle =C_R(a)$. We may set $K=\langle w\rangle\times \langle w'\rangle $, where $w^a=w'$, $w^2=u\in R- \langle z\rangle$, $(w')^2=uz$, $u^a=uz$, and $(w')^a=ws_0$ with $s_0\in R$. We have $s_0\neq 1$ (otherwise $a$ induces on $K$ an involutory automorphism). Since $$(ww')^2=w^2(w')^2=uuz=z , $$ it follows that $ww'\in R\langle v\rangle $. We compute $$(ww')^a=w'ws_0=(ww')s_0 , $$ which implies $\langle ww'\rangle \neq \langle v\rangle $ and so $\langle ww'u\rangle =\langle v\rangle $. Here we have used the fact that $R\langle v\rangle $ contains exactly two cyclic subgroups of order $4$ and they are $\langle v\rangle $ and $\langle vu\rangle$. This gives (since $a$ centralizes $v$) $$ww'u=(ww'u)^a=w'ws_0uz , $$ and consequently $s_0=z$. Hence $(w')^a=wz$ and so the action of $\langle a\rangle$ on $K$ is uniquely determined. We see that $$[a,w^{-1}]=w^aw^{-1}=w'w^{-1}=w'wu=ww'u $$ and so $G'= \langle ww'u\rangle$. But $\langle ww'u\rangle =\langle v\rangle\leq \langle a\rangle $ and so $\langle a\rangle$ is normal in $G$. Also, $\langle w\rangle $ induces on $\langle a\rangle$ an automorphism of order $4$ with $ \langle w\rangle\cap \langle a\rangle =1 $ and $|\langle a\rangle : C_{\langle a\rangle }(w) |=4$. This determines uniquely the structure of our "splitting" metacyclic group $$G=\langle a,w\,|\,a^{2^m}=w^4=1,\,\,m\geq 4,\,\, a^w=a^{1+2^{m-2}} \rangle , $$ where $H=\langle w, a^{2^{m-3}}\rangle $. For $m=4$, $H$ is nonabelian and for $m>4$, $H$ is abelian of type $(8,4)$. Here $G'=\langle a^{2^{m-2}} \rangle \leq Z(G)= \langle a^4\rangle$. We have to study the difficult case, where $G/K$ is generalized quaternion. Since a generalized quaternion group is not a subgroup of $Aut(K)$ (see Proposition 1.11 in Janko [3]), we have $C_G(K)\geq H$ and consequently $H$ is abelian of type $(8,4)$. Note that $H/K=Z(G/K)$. We set $L/K=(G/K)'=\Phi (G/K)$ so that $L/K$ is cyclic containing $H/K$. We have $G/L\cong E_4$ and since $G$ is metacyclic (and so $d(G)=2$), we get $L=\Phi (G)$ and $G'$ is cyclic. On the other hand, $G'$ covers $L/K$ and all elements in $H-K$ are of order $8$. Hence $G'\cap H\cong C_8$, $G'\cap K\cong C_4$, and $|L:G'|=4$. Again, since $G$ is metacyclic, there is a cyclic normal subgroup $S$ of $G$ such that $G'2^5$ with $|\Omega _3(G)|=2^5$. Then one of the following holds. \begin{itemize} \item[(a)] $G$ is abelian of type $(2^m,4),\,m\geq 4$. \item[(b)] $G$ is minimal nonabelian and more precisely $$G= \langle a,b\,|\,a^{2^m}=b^4=1,\,\,m\geq 4,\,\, a^b=a^{1+2^{m-1}}\,\,or \,\,b^a=b^{-1} \rangle . $$ \item[(c)] $$G= \langle a,l\,|\,a^{2^m}=l^8=1,\,\,m\geq 4,\,\,a^{2^{m-1}}=l^4,\,\, l^a=l^{-1} \rangle , $$ where $\Omega _3(G)= \langle l,\,a^{2^{m-3}} \rangle$ is abelian of type $(8,4)$, $G'=\langle l^2 \rangle\cong C_4 $, $Z(G)= \langle a^2 \rangle\cong C_{2^{m-1}} $, and $G'\not\leq Z(G)$. \item[(d)] $$G= \langle a,w\,|\,a^{2^m}=w^4=1,\,\,m\geq 4,\,\, a^w=a^{1+2^{m-2}} \rangle , $$ where $\Omega _3(G)=\langle w,\,a^{2^{m-3}} \rangle $, $G'=\langle a^{2^{m-2}} \rangle\cong C_4 $, $Z(G)=\langle a^4 \rangle $, and so $G'\leq Z(G)$. Also, for $m=4$, $\Omega _3(G)$ is nonabelian and for $m>4$, $\Omega _3(G)$ is abelian of type $(8,4)$. \item[(e)] $$G= \langle a,b\,|\,a^{2^m}=b^{16}=1,\,\,m\geq 4,\,\, a^{2^{m-1}}=b^8,\,\,a^b=a^{-1+\epsilon \cdot 2^{m-2}}, \,\,\epsilon =0,1 \rangle , $$ where $\Omega _3(G)=\langle b^2,\,a^{2^{m-3}} \rangle $ is abelian of type $(8,4)$ and $G/\Omega _2(G)\cong Q_{2^{m-1}}$. If $\epsilon =0$, $C_G(\Omega _2(G))$ is of index $2$ in $G$ and if $\epsilon =1$, then $C_G(\Omega _2(G))$ is of index $4$ in $G$. \end{itemize} \end{thm} \section{2-groups with exactly one subgroup of order $2^5$ and exponent $\leq 8$} $\,\,\,\,\,\,\,$ We prove here the following result. \begin{thm} Let $G$ be a 2-group containing exactly one subgroup of order $2^5$ and exponent $\leq 8$. Then we have $|\Omega _3(G)|=2^5$. \end{thm} In fact, with the similar proof, we get the following more general result. \begin{thm} Let $G$ be a 2-group containing exactly one subgroup of order $2^{n+2}$ and exponent $\leq 2^n$, where $n\geq 2$ is a fixed integer. Then we have $|\Omega _n(G)|=2^{n+2}$. \end{thm} \Prf Let $H$ be the unique subgroup of order $2^{n+2} $ and exponent $\leq 2^n\,\,(n\geq 2)$. Then $H$ is a characteristic subgroup of $G$ and $H\leq \Omega _n(G)$. Suppose that the theorem is false. Then there exists an element $a\in G-H$ of order $\leq 2^n$ with $a^2\in H$. Set $\tilde{H}= H\langle a \rangle $ so that $|\tilde{H}|=2^{n+3}$. Since $H$ is neither cyclic nor of maximal class, $H$ possesses a $G$-invariant 4-subgroup $R$. We have $|R\langle a \rangle |\leq 2^{n+2}$ and so there is a maximal subgroup $M$ of $\tilde{H}$ such that $M\geq R\langle a \rangle$. We have $exp(M\cap H)\leq 2^n$ and since $M=(M\cap H)\langle a \rangle$, we get $\Omega _n(M)=M $, $|M|=2^{n+2}$, $M\neq H$, and $exp(M)\leq 2^{n+1}$. The uniqueness of $H$ forces $exp(M)= 2^{n+1}$ and so $M$ has a cyclic subgroup of index $2$. But $|M|\geq 2^4$ (since $n\geq 2$) and $M$ possesses the normal 4-subgroup $R$. This implies that $M$ is not of maximal class and so $M$ is either abelian of type $(2^{n+1},2)$ or $M\cong M_{2^{n+2}}$. In both cases $\Omega _n(M)