Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
    
shot from the hip: let G be the group, A the subgroup of index 2. G acts on A, write A as a direct sum of the subgroup T on which G acts trivially, and the subgroup U on which G does not act trivially. Then show that G is the direct sum of a group isomorphic to T and the dihedral group over U. –  Franz Lemmermeyer Apr 15 '12 at 18:43
1  
Hmm, what about the quternion group. According to your consideration the quaternion group would be dihedral. –  Ralph Apr 15 '12 at 19:03
1  
@Franz: It does not necessarily split into a direct sum, unless the order of the group is odd. –  Will Sawin Apr 15 '12 at 19:33
    
As explained in my answer below, one can in essential restrict to 2-groups. I wonder, if it's possible to set up an induction there. In the well-known case of an elementary abelian maximal subgroup this seems possible. –  Ralph Apr 19 '12 at 15:53
add comment

1 Answer 1

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):

  • $G$ is a group with an abelian subgroup $A$ of index 2
  • $P$ is the Sylow 2-subgroup of $G$
  • $A=B \times D$ where $D$ is the Sylow 2-subgroup of $A$ and $|B|$ is odd.
  • $C_2=\langle t \rangle$ is the cyclic group of order 2.
  • $y \in G$ is choosen so that its coset generates $G/A$

Lemma:

  1. $B, D$ are normal in $G$ and $G = B \rtimes P$.
  2. $B$ is a $C_2$-module by $t \cdot b = yby^{-1}$.
  3. $y$ can be choosen such that $y^2 \in D$.

Theorem 1:

  1. If $G_1, G_2$ are isomorphic, then $P_1,P_2$ are isomorphic, and $B_1, B_2$ are isomorphic as $C_2$-modules.

  2. If $|[P_i,P_i]|> 2\;(i=1,2)$ then the converse is also true: If $P_1,P_2$ are isomorphic and $B_1,B_2$ are isomorphic as $C_2$-modules, then $G_1, G_2$ are isomorphic.


Theorem 1 boils the problem down to determine the $C_2$-modules $B$ and the groups $P$. This is covered by

Theorem 2:

  1. There is a bijection between the $C_2$-modules $B$ and the conjugacy classes of automorphisms $\varphi \in Aut(B)$ with $\varphi^2=id$.

  2. There is a bijection between the isomorphism classes of 2-groups $P$ with $|[P,P]|>2$
    having an abelian maximal subgroup and $$\coprod_\psi H^2(C_2,D_\psi)/C_{Aut(D)}(\psi)$$ where $\psi$ runs through (a system of representatives of) the conjugacy classes of $Aut(D)$ with $\psi^2=id$ such that $|\lbrace\psi(d)d^{-1}\mid d \in D\rbrace|>2$ (this is just the condition that makes $|[P,P]>2|$). Futhermore $D_\psi$ denotes the $C_2$-module with action $t \cdot d = \psi(d)$ and the action of the centralizer $C_{Aut(D)}(\psi)$ on the cohomology is induced by its action on $D$.


Example: Let use Theorem 2 to determine the isomorphism classes of 2-groups $P$ 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 non-split 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 non-abelian 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 2-subgroups $P_1,P_2$ 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 semi-direct 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
$$1 \to D \to P \to C_2 \to 1$$ and orbits of the action of a group $T$ and $H^2(C_2;D_\psi)$ is established. Since the only automorphism of $C_2$ is the identity, $T$ is just $C_{Aut(D)}(\psi)$. That there is a bijection can be proved similar to the proof of statement a) in the quoted link (replace $Z(G)$ by $D$) using that $D$ is the only abelian maximal subgroup of $P$.

share|improve this answer
2  
You need to make sure that the a_i generate your group A, otherwise A is not a normal subgroup of G. e.g. if you take the group <x,y | x^0 = 1, yxy^{-1} = x^2> then since the order of x^2 is the same as that of x, it seems to fit your description. But this group doesn't have an abelian subgroup of index 2. You also need to make sure that once the a_i generate A, the isomorphism given by x_i -> a_i gives an automorphism of A of order 2. Otherwise, if y^2x_iy^{-2} is not equal to x_i in A, you're imposing extra relations on A. You'd still have an abelian subgroup of index 2, just not A. –  Daniel Groves Apr 16 '12 at 0:16
    
Daniel, thanks for the hint. Sure you are right. This formulation should be correct: There is an automorphism $\varphi$ of order 2 of $A$ such that $yx_iy^{-1}=\varphi(x_i)=y^{-1}x_iy$. The condition on the exponents can be omitted then. –  Ralph Apr 16 '12 at 0:41
    
Metin Atalin says (in a deleted answer) "Thanks, this is helpful. I was thinking of finite groups G with abelian subgroups of index 2." –  S. Carnahan Apr 17 '12 at 9:17
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.