From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: conjugate classes in a group Date: 22 Jul 1998 18:16:08 GMT In article <01bdb3c0$becadae0$37b939c2@buromath.ups-albi.fr>, ortiz wrote: >In a finite group, conjugate elements have the same order. >I try to find all groups where the converse is true ie finite groups such >that: > > same order elements are in same conjugate classe. > >For example, S_3 has 3 conjugate classes and no elements >in distincts classes have same order. Well! This certainly seemed like the sort of thing someone somewhere would have investigated, but it took a little sleuthing to find this reference in Math Reviews: 87b:20021 20D05 Fitzpatrick, Patrick(IRL-CORK) Order conjugacy in finite groups. (English) Proc. Roy. Irish Acad. Sect. A 85 (1985), no. 1, 53--58. _________________________________________________________________ Let $\sigma$ be a set of primes. For a finite group $G$, define $G\sb n$ to be the set of all elements of $G$ of order $n$. $G$ is said to be $\sigma$-conjugate if all elements of $G\sb n$ are conjugate in $G$ whenever $n$ is a $\sigma$-number. If $\pi(G)$ is the set of all primes dividing the order of $G$, then $G$ is said to be order-conjugate. A cyclic group of order 2 and the symmetric group of degree 3 are order-conjugate. It is shown in the paper under review that these two groups are the only order-conjugate groups. The property "$\sigma$-conjugate" is inherited by factor groups, which is a little surprising but easy to prove. Hence it suffices to show that the group of order 2 is the only simple order-conjugate group and the only possible normal subgroup is of order 3. The proof depends on the classification of all simple groups of finite order. Reviewed by Koichiro Harada Cited in: 97h:20018 95d:20029 90a:20016 87k:20031 So the given example is the most complicated one there is, and yet the proof requires quite a bit of machinery! dave ============================================================================== From: ikastan@sol.uucp (ilias kastanas 08-14-90) Newsgroups: sci.math Subject: Re: conjugate classes in a group Date: 23 Jul 1998 01:50:28 GMT In article <01bdb3c0$becadae0$37b939c2@buromath.ups-albi.fr>, ortiz wrote: @In a finite group, conjugate elements have the same order. @I try to find all groups where the converse is true ie finite groups such @that: @ @ same order elements are in same conjugate classe. For any G there is an H, G subgroup of H and H having this property. It's a theorem due to HNN. However, the H constructed by the proof is uncountable... @For example, S_3 has 3 conjugate classes and no elements @in distincts classes have same order. @ @Obviously, no abelian finite group with order >2 have this property. Just as obviously, no S_n with n > 3 will do either ( e.g. take (a b), (a b)(c d) ). In fact, I don't think any other finite group at all can have the property. But proving so will probably be messy. Ilias ============================================================================== Date: Fri, 24 Jul 98 4:20 +0300 To: rusin@vesuvius.math.niu.edu Subject: Re: conjugate classes in a group From: mann@vms.huji.ac.il (Avinoam Mann) Newsgroups: sci.math Subject: Re: conjugate classes in a group In article <6p5a98$54h$1@gannett.math.niu.edu>, rusin@vesuvius.math.niu.edu (Dave Rusin) writes... >In article <01bdb3c0$becadae0$37b939c2@buromath.ups-albi.fr>, >ortiz wrote: >>In a finite group, conjugate elements have the same order. >>I try to find all groups where the converse is true ie finite groups such >>that: >> >> same order elements are in same conjugate classe. >> >>For example, S_3 has 3 conjugate classes and no elements >>in distincts classes have same order. > >Well! This certainly seemed like the sort of thing someone somewhere would >have investigated, but it took a little sleuthing to find this reference >in Math Reviews: This problem was solved in three different papers published at more or less the same time. Besides Fitzpatrick's paper below, there is a paper by R.v.d. Waall, and a collaborator whose name escapes me, I think in Indag. Math., and a paper by W.Feit and G.Seitz in Ill. J. Math., which goes considerably further. Their formulationof the result is: the only finite groups with this conjugacy property are the symmetric groups S(n) for n at most 3. Avinoam Mann [Rusin article quoted, now omitted -- djr]