Hi,
Is there a general presentation for a group with an abelian subgroup of index 2? Is there a classification of such groups.
Thanks
Metin Analin
Hi, Is there a general presentation for a group with an abelian subgroup of index 2? Is there a classification of such groups. Thanks Metin Analin 


Let $A$ be the abelian subgroup of index 2. If $A$ is finitely generated, choose a presentation $$A = \langle x_1,...,x_n\mid \forall i,j:\; [x_i,x_j]=1, x_i^{e_i} = 1 \rangle$$ where $e_i = 0$ if $x_i$ is not of finite exponent. Then the groups $G$ having $A$ as index 2 subgroups are exactly the groups with the presentation $$G=\langle x_1,...,x_n,y\mid \forall i,j: [x_i,x_j]=1,x_i^{e_i}=1,y^2=a,yx_iy^{1}=\varphi(x_i)\rangle$$ where $a \in A$ and $\varphi(x_i) \in Aut(A)$ with $\varphi^2=1$. Since the OP is interested in finite groups, I'll specialize now to the finite case. The following notation is used (sometimes also with subscripts):
Lemma:
Theorem 1:
Theorem 1 boils the problem down to determine the $C_2$modules $B$ and the groups $P$. This is covered by Theorem 2:
Example: Let use Theorem 2 to determine the isomorphism classes of 2groups $P$ having an elementary abelian subgroup $D=(\mathbb{Z}/2)^n$ of index 2. By linear algebra, the only elements of $Aut(D)=GL_n(\mathbb{F}_2)$ of order $\le 2$ are (up to conjugacy) the matrices $\psi=\begin{pmatrix}I_r & 0 \newline 0 & J\end{pmatrix}$ where $J$ is a block diagonal matrix with, say, $s$ Jordan blocks $\begin{pmatrix}1 & 1 \newline 0 & 1\end{pmatrix}$. $D^{C_2}$ is just the eigenspace of $\psi$ and we have $H^2(C_2;D_\psi)\cong \mathbb{F}_2^r$ (the eigenvectors from the Jordan blocks get killed). Futhermore, the centalizer $C(\psi)$ includes $\tilde{A} =\begin{pmatrix}A & 0 \newline 0 & I\end{pmatrix}$ where $A$ runs through $GL_r(\mathbb{F}_2)$. The induced action on the cohomology is given by $\mathbb{F}_2^r \to \mathbb{F}_2^r,\; x \mapsto Ax$ and is hence transitive (if $x\neq 0$). Hence for each $\psi$ there are two orbits (one split extension and a nonsplit one). Eventually, the (somewhat obscure) condition that $(I+\psi)\mathbb{F}_2> 2$ is equivalent to $\psi$ having to least two Jordan blocks. So Theorem 2 shows that there are exactly $$2\lbrace 2 \le s \le n/2\rbrace = 2\lfloor n/2 \rfloor 2$$ isomorphism classes of $P$'s with more than two commutators. In particular, in case $n=2$ we see that there is no such group (what is right, since the only nonabelian group of order 8 with an elementary abelian max. subgroup is the dihedral group whose commutator subgroup has order 2). Remark: By the previous results, the isomorphism classes of the groups $G$ is completely determined by the conjugacy classes of automorphisms of order 2 of the abelian groups $B$, $C$ and its centralizers. A description of the automorphism group of finite abelian groups can be found in this paper. Proof of the Lemma: Up to 2) it's an easy excercise. By writing $y^2=bd$ we see $bd=(yby^{1})(ydy^{1})$, i.e. $b=yby^{1}=t \cdot y$ is invariant under the $C_2$action. Since $B$ is odd, $H^2(C_2,B)=0=B^{C_2} /(1+t)B$. Hence there is $c \in B$ such that $b^{1}=(1+t)c=c(ycy^{1})$. Now $(yc)^2=d$ and replacing $y$ by $yc$ does the trick. Proof of Theorem 1: 1) Let $\phi: G_1 \to G_2$ be an isomrphism. Then the Sylow 2subgroups $P_1,P_2$ are also isomorphic and $B_1=B_2$ follows. Let $b \in B_1$ and $\phi(b)=ay^i$ with $a \in A_2$ and $y=y_2$. Let $m$ be odd with $b^m=1$. Then $\bar{y}^{mi}=\bar{y}^i=\bar{1}$ in $G_2/A_2$. Thus $i$ is even and $\phi(b) \in A_2$. By an order argument, $\phi(b) \in B_2$, i.e. $\phi(B_1) \subseteq B_2$ and because $\phi$ is injective and $B_1=B_2$, we have $\phi(B_1)=B_2$. It remains to show $B_1 \cong B_2$ as $C_2$modules, i.e. we have to show $$\phi(y_1by_1^{1})=y_2\phi(b)y_2^{1},\quad\quad b \in B_1\hspace{50pt}(\ast)$$ Write $\phi(y_1)=ay_2^i$ with $a\in A_2$. If $i$ were even, then $y_2^i \in A_2$ and $\phi(G_1) \subseteq A_2 \varsubsetneqq G_2$. Hence $i$ is odd we may assume $i=1$. Thus $$\phi(y_1by_1^{1})= \phi(y_1)\phi(b)\phi(y_1)^{1}=a(y_2\phi(b)y_2^{1})a^{1}=y_2\phi(b)y_2^{1}$$ (the $a$ cancels since we know $\phi(b) \in B_2$ which is normal in $G_2$), proving $(\ast)$. 2) Let $\varphi: P_1 \to P_2$ be an isomprphism and let $\phi: B_1 \to B_2$ be an isomorphism of $C_2$modules. If the following diagramm commutes, then the semidirect products $G_i = B_i \rtimes P_i$ are isomorphic: $$\begin{array}{ccc} P_1 & \xrightarrow[]{\beta_1} & Aut(B_1) \newline \varphi\downarrow\; & a & \downarrow\phi^\ast \newline P_2 & \xrightarrow[\beta_2]{} & Aut(B_2) \end{array}$$ Let $g=dy_1^i \in P_1$ and $b \in B_2$. Then an easy computation using $(\ast)$ gives $$\phi^\ast(\beta_1(g))(b) = y_2^iby_2^{i}.$$ Now assume the commutator subgroup of $P_i$ has more than two elements. By using for example Lemma 4.6 of Isaacs: Finite Group Theory, it's not hard to see that $D_i$ is the only abelian maximal subgroup of $P_i$. Since the isomorphism $\varphi$ preserves these properties, $\varphi(D_1)=D_2$ follows. Hence $\varphi(d)=e \in D_2$. Write $\varphi(y_1)=fy_2^j,\;f \in D_2$. Since $j$ is odd (otherwise $\varphi(P_1) \subseteq D_2 \varsubsetneqq P_2$) and $y_2^2 \in D_2$ we may assume $j=1$. A simple induction shows $\varphi(y_1^i)=(fy_2)^i =hy_2^i$ for some $h \in D_2$. Thus $$\beta_2(\varphi(g))(b)=\varphi(g)b\varphi(g)^{1}=eh(y_2^i b y_2^{i})h^{1}e^{1}=y_2^i b y_2^{i}$$ (the latter holds since $B_2$ is normal in $G_2$) and the diagramm commutes. Proof of Theorem 2: 1) is obvious. 2) Fix $\psi \in Aut(D)$. In Counting isomorphism classes via extensions a bijection between weakly equivalent extensions 

