Talk:Holomorph (mathematics)

Example

I don't feel confident enough to write it, but I think an example would enhance this page (which is already very good). E.g. Hol(Z_3) is isomorphic to D_6 Mathematical Leopard 01:10, 27 September 2006 (UTC)[reply]

Origin of name?

Where does the name come from? —vivacissamamente 13:00, 4 January 2007 (UTC)[reply]

Example and generalization

I think Hol(Z_2 x Z_2) being isomorphic to S_4 is also a good example.

Also, something strange was recently brought to my attention by playing around: Iff G is nonabelian, antiautomorphisms are not automorphisms. Adjoining antiautomorphisms to the holomorph, in this case, doubles the order of the group. The stabilizer of the identity in this extended group is now isomorphic to Aut(G) x Z_2, since one antiautomorphism, inversion in G, commutes with all automorphisms of G.

Some interesting examples of this larger group: If G=S_3, then this larger group is permutation-isomorphic to the wreath product S_3 wr S_2; in terms of S_3, the blocks for this action are A_3 and S_3 \ A_3. In other words, the only thing preserved by the action of this larger group on S_3 is whether or not two permutations have the same sign.

If G is the dihedral group of order 8, then this larger group is a Sylow 2-subgroup of S_8, acting on the elements of the dihedral group.

If G is the quaternion group Q_8 of order 8, then this larger group is permutation-isomorphic to the wreath product S_2 wr S_4; in terms of Q_8, the blocks for this action are the cosets of {+/- 1}. In other words, the only thing preserved by the action of this larger group on Q_8 is whether or not two elements differ only in sign.

This larger group is also describable as the join of the holomorphs of the left and right regular representations of G.

A question: Is this larger group always the normalizer of Hol(G) in S_|G| (acting on the elements of G) when G is nonabelian?

67.100.30.124 (talk) 06:06, 27 February 2008 (UTC)I don't yet have a Wikipedia username.[reply]

Thanks! Definitely S4=Hol(2 x 2) needs to be added. Your description of the larger group is quite interesting, but I believe there is a small mistake just for holomorphs.

For the symmetric group on three points, the quaternion group of order 8, or the dihedral group of order 8: for the size of the group n, the regular representation L, the centralizer R of L in S8, the normalizer H of L in Sn, the normalizer K of R in Sn, and the normalizer A of H in Sn, one has H=K, [A:H]=2. In other words, the left and right holomorphs coincide.

The calculations of the larger group appear correct otherwise though. Let the larger group be called Ant(G). Ant(D8) = 2 wr 2 wr 2. Ant(Q8) = 2 wr S4, Ant(S3) = S3 wr 2. In each case it is the full normalizer. I also checked G=A5.

I don't think it matters whether the regular representations are both taken as multiplication, or multiplication by the inverse.

Ant(G) is often the normalizer of H in Sn, but not always. For instance, not for the groups of maximal class and order 16 (D16, SD16, Q16), and not for some groups of order 24, including C4 x S3 and D24.

Let me know if you find references for these ideas, it would be wonderful to include them. JackSchmidt (talk) 06:59, 27 February 2008 (UTC)[reply]

Note a sci.math post by David Harden. JackSchmidt (talk) 06:00, 9 March 2008 (UTC)[reply]

I noted that post already because I am its poster. Thanks for the correction; the left and right holomorphs always coincide for nonabelian G, since the centralizer of a group is contained in its normalizer and the left and right regular representations of G are S_|G|-centralizers of each other. 67.103.203.10 (talk) 07:40, 10 March 2008 (UTC)I intend to get an accessible username soon. (I had one, and then I forgot the password due to disuse)[reply]

Another example, please

It would be good to have an example of the holomorph of a non-Abelian group.

The article says, fairly clearly, that the holomorph of G is the semidirect product of G by Aut(G). A natural mistake is to think that the holomorph of G is the semidirect product of G by (Aut(G)/Inn(G)). An example where G in not Abelian would make this clearer. Maproom (talk) 23:18, 14 July 2011 (UTC)[reply]

An example of the application of holomorph

I think it is worth mentioning that: a irreducible polynomial of degree $p$ (a prime) is solvable in radical if and only if all roots of the polynomial can be expressed as rational functions of its any two roots given, and if and only if its Galois group is a subgroup of the holomorph of the cyclic group of order $p$ (see here). I believe that this is a famous theorem established in the 19th century. 129.104.241.231 (talk) 08:34, 26 September 2024 (UTC)[reply]