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I'm trying to work through a sketch proof attributed to Walter Feit on characterizing $S_5$.

Suppose $G$ is a finite group with exactly two conjugacy classes of involutions, with $u_1$ and $u_2$ being representatives. Suppose $C_1=C(u_1)\simeq \langle u_1\rangle\times S_3$ and $C_2=C(u_2)$ be a dihedral group of order $8$. The eventual result is that $G\simeq S_5$. Also, $C(u)$ denotes the centralizer of $u$ in $G$.

I'm trying to deduce that $|S_2|=0$ or $4$, where $S_i$ is the set of pairs $(x,y)$ with $x$ conjugate to $u_1$, $y$ conjugate to $u_2$, and $(xy)^n=u_i$ for some $n$, and that $C_2$ then has a noncyclic subgroup $V$ such that all involutions in $V$ are conjugate to $u_2$ in $G$. This is part of exercise 10 of page 83, Jacobson's Basic Algebra I. Thanks for any help.


My sparse thoughts: I know a few facts I think are useful (proofs are linked in the numbers to the left hand side):

1. If $c_i=|C(u_i)|$ and $s_i=|S_i|$, then $|G|=c_1s_2+c_2s_1$.

2. $C_2$ is a Sylow $2$-subgroup, and I observe from this that one can assume $u_1\in C_2$ by taking a conjugate.

3. There are $3$ classes of involutions in $C_2$, and if $x$ is an involution distinct from $u_2$, then $x$ is conjugate to $xu_2$ in $C_2$.

I'm lost on how to use this info on the involutions in $C_2$ to count the size of $S_2$. Identifying $C_2$ with $D_8=\langle r,s\mid r^4=s^2=1,\; sr=r^3s\rangle$ and $r^2$ with $u_2$, I know that the two noncyclic subgroups of order $4$ are $\{1,s,r^2,r^2s\}$ and $\{1,rs,r^2,r^3s\}$. I consider the intersection $C:=C_1\cap C_2$. This is a subgroup with order dividing $8$ and $12$, but since I already know $1,u_1,u_2\in C$, then $|C|=4$. I think this might be the desired subgroup $V$. I'm not sure what action of $C_1$ on $u_2$ to consider, since I don't know if left multiplication or conjugation will send $u_2$ back into $C_2$.

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Consider the intersection of C1 and C2 and how C1 acts on u2. –  Jack Schmidt Jun 6 '12 at 16:31
    
Thanks @JackSchmidt, I could deduce that $C_1\cap C_2$ has order $4$, but I don't know what sensible action to look at. How would I know if left multiplication or conjugation would send $u_2$ into $C_2$? –  Adelaide Dokras Jun 7 '12 at 1:22
    
My apologies. C1 does not stabilize C, and C is the wrong subgroup. I need to use a small result on fusion (Frobenius or Burnside), and then the centralizers don't matter. If G is a finite group with a dihedral Sylow 2-subgroup P and G has exactly 2 conjugacy classes of involutions, then P has two P-conjugacy classes of non-cyclic subgroups of order 4, and exactly one of those classes has the property that all of its involutions are G-conjugate. However, I need to use the result that conjugacy is determined by normalizers of subgroups of P (which here follows from Frobenius or Burnside). –  Jack Schmidt Jun 7 '12 at 6:34
    
@JackSchmidt Thanks, but isn't the majority of that result what I'm trying to prove? –  Adelaide Dokras Jun 7 '12 at 7:05
    
Yes. You are just trying to describe the fusion in C2, which is easy to do if you use any results on fusion, like Burnside or Frobenius. –  Jack Schmidt Jun 7 '12 at 7:24
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2 Answers 2

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As noted in the OP, we may assume $u_1\in C(u_2)$, i.e. $u_1$ and $u_2$ commute.

The key property here is what is called "Exercise 8" in Jacobson's book : in a dihedral group of even rank, the largest element is central.

This immediately implies that $x$ and $y$ are both in $C(u_i)=C_i$ whenever $(x,y) \in S_i$. Denote by $I_2$ the set of involutions in $C_2$ other than $u_2$, $X_2$ the set of involutions in $I_2$ that are conjugate (in $G$) to $u_1$, and $Y_2$ the set of involutions in $I_2$ that are conjugate (in $G$) to $u_2$. Then we must count the elements in $$ S_2= X_2\times Y_2 $$

In $I_2$ there are two $C_2$-conjugacy classes of cardinality $2$. If those two classes merge in $G$, then $X_2=I_2$ and $Y_2=\emptyset$, so $s_2=0$. If those two classes do not merge, then $|X_2|=|Y_2|=2$ so $s_2=2 \times 2=4$.

We do something similar for $S_1$. There is a subgroup $A$, isomorphic to ${\mathfrak S}_3$ (let us denote its elements by $a_{\sigma}, \sigma \in {\mathfrak S}_3$), such that $C_1$ is the disjoint union of $A$ and $u_1A$. Note that we may replace $A$ with $A'=\lbrace u_1^{{\sf signature}(\sigma)}a_{\sigma}| \sigma \in {\mathfrak S}_3 \rbrace$ which is a subgroup sharing exactly the same properties as $A$. If $u_2\not\in u_1A$ then $u_2\in u_1A'$, so we may assume without loss of generality that $u_2\in u_1A$ : $u_2=u_1a_{\tau}$ for some $\tau$ which must be of order $2$. We can assume $\tau=(1,2)$.

Denote by $I_1$ the set of involutions in $C_1$ . Also, put $$ \Gamma=\lbrace (x,y) \in I_1^2| \exists n, (xy)^n=u_1 \rbrace $$

There are three conjugacy classes in $I_1$ :

$$ K_1=\lbrace u_1 \rbrace, \ K_2=\lbrace u_1a_{(i,j)} | 1 \leq i \lt j \leq 3 \rbrace, \ K_3=\lbrace a_{(i,j)} | 1 \leq i \lt j \leq 3 \rbrace $$

It is easy to check that $$ \Gamma = (K_2 \times K_3) \cup (K_3 \times K_2) $$

Now $u_1$ and $u_1u_2$ are conjugates in the dihedral group $C_2$. We deduce that $K_3$ merges with $K_1$ in $G$, so $S_1=K_3 \times K_2$ so $s_1=9$.

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I hate mathjax. Why does it "swallow" and partly hide the paragraph starting with "This immediately ..."? –  Ewan Delanoy Jun 19 '12 at 14:06
    
that is a "markdown" problem. You cannot start a line with 4 spaces. –  Jack Schmidt Jun 19 '12 at 14:07
    
@JackSchmidt :thanks. –  Ewan Delanoy Jun 19 '12 at 14:12
    
No problem. Thanks for posting a solution. I still haven't had a chance to look through Jacobson for this exercise. –  Jack Schmidt Jun 19 '12 at 14:13
    
Thank you Ewan!, I did not notice you had posted an answer until now. –  Adelaide Dokras Jun 24 '12 at 0:29
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Let G be a finite group with Sylow 2-subgroup P, and suppose G has exactly two conjugacy classes of involutions and P is dihedral of order eight, then we use Sylow theorems to determine the "fusion pattern" of the involutions of P.

Proposition: If G has exactly two conjugacy classes of involutions and P is a dihedral Sylow 2-subgroup of order eight, then P has "PGL" type fusion, that is, P has one non-cyclic subgroup of order 4 in which all involutions are G-conjugate and one non-cyclic subgroup of order 4 in which there are two G-conjugacy classes of involutions.

Proof: Let r in P have order 4, and let s be a non-central involution of P [as in the question]. Let V be non-cyclic abelian subgroup of P, say { 1, rr, s, rrs }, and let W be the other non-cyclic abelian subgroup of P, say { 1, rr, rs, sr }. Note that s and rrs are already P-conjugate (by r), and rs and sr are already P-conjugate (again by r). Even though P has three conjugacy classes of involutions, K1 = { rr }, K2 = { s, rrs }, and K3 = { sr, rs }, G has only two, so two of these conjugacy classes "fuse" in G.

One possibility is K1 and K2 fuse, giving V as the required subgroup. Another possibility is that K1 and K3 fuse, giving W as the required subgroup. It is not possible that K2 and K3 fuse. In general, this follows from Frobenius's normal p-complement theorem (or most other early results on fusion), but I'll use an argument of Grün to make it simpler in your case. By way of contradiction, suppose $(rs)^g = s$ for some g in G.

Then $s \in P \cap P^g$ so s commutes both with $Z(P)$ and $Z(P)^g$, and so $Z(P), Z(P)^g \leq C_1:=C_G(s)$. By Sylow's theorem, some $C_1$-conjugate of $Z(P)$ is contained in the same Sylow 2-subgroup of $C_1$ that contains $Z(P)^g$, so that $Z(P)^c,Z(P)^g$ are both contained in some G-conjugate of P, say $Z(P)^{ch}, Z(P)^{gh} \leq P$ for some h in G. However, $Z(P) = \{ 1, r^2 \}$ and since K1 does not fuse with K2 or K3, the only possibility is $r^2 = (r^2)^{ch} = (r^2)^{gh}$. In particular, $gc^{-1} \in C_2 := C_G(r^2)$ and so $(rs)^{gc^{-1}} = s^{c^{-1}} = s$ so that K2 and K3 fuse already in $C_2$. However, in your case $C_2 = P$, so that s and rs are clearly not $C_2$-conjugate, a contradiction. $\square$

I'd be interested to see an answer that made use of $S_2$.

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Dear Jack, thanks very much for writing a neat answer at a level understandable for me. I too am interested in seeing how $S_2$ is used. –  Adelaide Dokras Jun 7 '12 at 20:21
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