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Noah wrote:

Here's an elementary observation, in the definition of non-splitting you can restrict your attention to those H's generated by a two (not necessarily distinct) elements of c.

As FC points out two involutions always generate a dihedral group. As JSE points out (G,c) for c a class of involutions is nonsplitting iff the resulting dihedral groups are all of 2*odd order. In particular, (G,c) is nonplitting for c an involution iff the product of any two elements of c has odd order.

JSE suggests looking at the subgroup generated by all pairwise products of elements in c. This either generates the whole group or generates an index 2 subgroup (which is necessarily normal and has a complement generated by any element of c). FC noted that instead of looking at pairwise products you could instead look at pairwise commutants (since the commutant of two involutions is just the square of their product and in an odd order cyclic group the square of a generator generates).

Let's concentrate on the latter case.

when does a group A admit an involution i:A-->A such that i(a)a^-1 always has odd order, and {i(a)a^-1} generates for all A generate A.

[I think the only such groups have odd order. Also, Since A has an odd number of 2-Sylow subgroups, i preserves at least one Sylow P. Yet then i(a)a^-1 lies in P, and is thus trivial. Thus i fixes P. It follows that i preserves the normalizer N of P.--FC

I'd just run through the same argument myself before realizing this is just the fact that the centralizer of an element of c in the big group contains the 2-Sylow and its normalizer (as in the Frattini answer).--Noah]

Does this have anything to do with H^1(Z/2, A)? To flesh this out, the maps Z/2->A sending the nontrivial element to i(a)a^-1 are exactly the coboundaries. --Noah

Wait a sec, since every coboundary is a cocycle (or by a direct one-line computation) if y = i(a) a^-1 then i(y) = y^-1. So in particular we'd need that A is generated by elements such that i(y) = y^-1. --Noah

Another characterization that (G,c) splits for c an involution (and <c> generates G) is that:

(i) G is generated by <c>, (ii) [g,c] has odd order for every g in G.

(the latter just says that the product (gcg^-1*c) of any two conjugates of c has odd order.) These conditions are preserved under taking quotients. Thus they hold for at least one simple group. Using the classification (urgh) I think from this one can deduce that the only simple quotient of G is Z/2Z. This would reduce the problem to the "first case" considered above. --FC.

Actually running through all involutions in all the simple groups sounds very hard to me. Is there some reason to expect that to be tractable? In particular, for the groups of Lie type? --Noah

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Noah wrote:

Here's an elementary observation, in the definition of non-splitting you can restrict your attention to those H's generated by a two (not necessarily distinct) elements of c.

As FC points out two involutions always generate a dihedral group. As JSE points out (G,c) for c a class of involutions is nonsplitting iff the resulting dihedral groups are all of 2*odd order. In particular, (G,c) is nonplitting for c an involution iff the product of any two elements of c has odd order.

JSE suggests looking at the subgroup generated by all pairwise products of elements in c. This either generates the whole group or generates an index 2 subgroup (which is necessarily normal and has a complement generated by any element of c). FC noted that instead of looking at pairwise products you could instead look at pairwise commutants (since the commutant of two involutions is just the square of their product and in an odd order cyclic group the square of a generator generates).

Let's concentrate on the latter case.

when does a group A admit an involution i:A-->A such that i(a)a^-1 always has odd order, and {i(a)a^-1} generates for all A generate A.

[I think the only such groups have odd order. Also, Since A has an odd number of 2-Sylow subgroups, i preserves at least one Sylow P. Yet then i(a)a^-1 lies in P, and is thus trivial. Thus i fixes P. It follows that i preserves the normalizer N of P.--FC

I'd just run through the same argument myself before realizing this is just the fact that the centralizer of an element of c in the big group contains the 2-Sylow and its normalizer (as in the Frattini answer).--Noah]

Does this have anything to do with H^1(Z/2, A)? To flesh this out, the maps Z/2->A sending the nontrivial element to i(a)a^-1 are exactly the coboundaries. --Noah

Another characterization that (G,c) splits for c an involution (and <c> generates G) is that:

(i) G is generated by <c>, (ii) [g,c] has odd order for every g in G.

(the latter just says that the product (gcg^-1*c) of any two conjugates of c has odd order.) These conditions are preserved under taking quotients. Thus they hold for at least one simple group. Using the classification (urgh) I think from this one can deduce that the only simple quotient of G is Z/2Z. This would reduce the problem to the "first case" considered above. --FC.

Actually running through all involutions in all the simple groups sounds very hard to me. Is there some reason to expect that to be tractable? In particular, for the groups of Lie type? --Noah

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Noah wrote:

Here's an elementary observation, in the definition of non-splitting you can restrict your attention to those H's generated by a two (not necessarily distinct) elements of c.

As FC points out two involutions always generate a dihedral group. As JSE points out (G,c) for c a class of involutions is nonsplitting iff the resulting dihedral groups are all of 2*odd order. In particular, (G,c) is nonplitting for c an involution iff the product of any two elements of c has odd order.

JSE suggests looking at the subgroup generated by all pairwise products of elements in c. This either generates the whole group or generates an index 2 subgroup (which is necessarily normal and has a complement generated by any element of c). FC noted that instead of looking at pairwise products you could instead look at pairwise commutants (since the commutant of two involutions is just the square of their product and in an odd order cyclic group the square of a generator generates).

Let's concentrate on the latter case.

when does a group A admit an involution i:A-->A such that i(a)a^-1 always has odd order, and {i(a)a^-1} generates for all A generate A.

[I think the only such groups have odd order. Also, Since A has an odd number of 2-Sylow subgroups, i preserves at least one Sylow P. Yet then i(a)a^-1 lies in P, and is thus trivial. Thus i fixes P. It follows that i preserves the normalizer N of P.--FC

I'd just run through the same argument myself before realizing this is just the fact that the centralizer of an element of c in the big group contains the 2-Sylow and its normalizer (as in the Frattini answer).--Noah]

Does this have anything to do with H^1(Z/2, A)? To flesh this out, the maps Z/2->A sending the nontrivial element to i(a)a^-1 are exactly the coboundaries. --Noah

Another characterization that (G,c) splits for c an involution (and <c> generates G) is that:

(i) G is generated by <c>`, (ii) [g,c] has odd order for every g in G.

(the latter just says that the product (gcg^-1*c) of any two conjugates of c has odd order.) These conditions are preserved under taking quotients. Thus they hold for at least one simple group. Using the classification (urgh) I think from this one can deduce that the only simple quotient of G is Z/2Z. This would reduce the problem to the "first case" considered above. --FC.

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