4
$\begingroup$

Which numbers $n$ have the property that every group with order $n$ has a non-trivial center ?

$n$ has this property if it is an abelian number (every group with order $n$ is abelian).

If $n$ is a prime-power, it also has the property.

If $n$ is even, but not divisible by $4$, $n$ does not have the property because the dihedral group of order $n$ has trivial center. And if $n=pq$ with primes $p<q$ and $p|q-1$, $n$ does not have the property.

But how can I check the property in general, for example for $n=28$ ? Does the class equation help ?

The numbers upto $511$, which do NOT have the property, are :

6   10   12   14   18   20   21   22   24   26
30   34   36   38   39   42   46   48   50   52
54   55   56   57   58   60   62   66   68   70
72   74   75   78   80   82   84   86   90   93
94   96   98   100   102   106   108   110   111   114
116   118   120   122   126   129   130   132   134   136
138   140   142   144   146   147   148   150   154   155
156   158   160   162   164   166   168   170   171   174
178   180   182   183   186   190   192   194   196   198
200   201   202   203   204   205   206   210   212   214
216   218   219   220   222   226   228   230   234   237
238   240   242   244   246   250   252   253   254   258
260   262   264   266   270   272   273   274   276   278
280   282   286   288   290   291   292   294   298   300
301   302   305   306   308   309   310   312   314   318
320   322   324   326   327   328   330   333   334   336
338   340   342   346   348   350   351   354   355   356
358   360   362   363   364   366   370   372   374   378
380   381   382   384   386   388   390   392   394   396
398   399   400   402   404   405   406   408   410   414
417   418   420   422   426   430   432   434   436   438
440   441   442   444   446   448   450   452   453   454
456   458   460   462   465   466   468   470   471   474
476   478   480   482   484   486   489   490   492   494
497   498   500   502   504   505   506   507   510
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  • 3
    $\begingroup$ If $n = 2^rm$ with $m$ odd and $m$ has at least $r$ factors, then you can take a direct product of $r$ dihedral groups. $\endgroup$ – Tobias Kildetoft Mar 23 '16 at 8:16
  • 2
    $\begingroup$ Or more generally, if you can factor $n$ into a product of numbers without the property, then $n$ also does not have the property. $\endgroup$ – Tobias Kildetoft Mar 23 '16 at 8:18
  • 2
    $\begingroup$ Hmm, it might be better to look for those $n$ for which there exists a group with trivial center of that order (i.e. reversing which property we want). Then note that the set of such $n$ is closed under multiplication. $\endgroup$ – Tobias Kildetoft Mar 23 '16 at 8:19
  • 1
    $\begingroup$ Don't know about the general case, but a group of order $28$ surely has a non-trivial center. Its Sylow-7 $P\simeq C_7$ is unique, hence normal. But $Aut(C_7)$ has only a single element of order two and none of order four, so some element $x$ of a Sylow-2 will necessarily commute with all of $P$, and hence be in the center (Sylow-2 is abelian, so the centralizer of $x$ has to be the entire group). $\endgroup$ – Jyrki Lahtonen Mar 23 '16 at 8:26
  • 3
    $\begingroup$ All nilpotent numbers have this property. (See jstor.org/stable/2589118?seq=1#page_scan_tab_contents and oeis.org/A056867) This generalises both your abelian number and prime power example. The converse is not true, 28 being the smallest counterexample. $\endgroup$ – verret Mar 23 '16 at 9:34

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