Conversation started Mar 10, 2012 at 2:28.
Kannappan Sampath
Kannappan Sampath
Mar 10, 2012 02:28
@Mariano Can you assist me with GAP please?
I setup GAP but I just cannot query anything into it, it replies with some error message.
So, I'd like it if you can query your GAP system for those groups G such that G and Aut G are abelian?
Mariano Suárez-Alvarez
Mariano Suárez-Alvarez
/me tries
Kannappan Sampath
Kannappan Sampath
@MarianoSuárezAlvarez Thanks for trying. : )
Mariano Suárez-Alvarez
Mariano Suárez-Alvarez
There are 200 of size at most 200
The cyclic ones :)
Kannappan Sampath
Kannappan Sampath
@MarianoSuárezAlvarez Oh, so, the order of the groups are bounded by 200?
Mariano Suárez-Alvarez
Mariano Suárez-Alvarez
No, I limited the search at 200
I guess the only such groups are the cyclic groups
Kannappan Sampath
Kannappan Sampath
Mar 10, 2012 02:36
@MarianoSuárezAlvarez I should attempt a proof. This question was asked in a paper in American Math. Monthly.
Mariano Suárez-Alvarez
Mariano Suárez-Alvarez
a group G satisfies your conditions iff its primary components satisfy them
so it is enough to suppose that |G|=p^r for some prime p
Kannappan Sampath
Kannappan Sampath
(In fact, IRC they said this was not interesting or something. : ) )
Mariano Suárez-Alvarez
Mariano Suárez-Alvarez
then the group is a direct sum of cyclic groups of the form Z_{p^k}
Kannappan Sampath
Kannappan Sampath
@MarianoSuárezAlvarez I am not getting this point. Can you explain why so, please?
Mariano Suárez-Alvarez
Mariano Suárez-Alvarez
if there are two isomorphic sumands, then Aut(G) is not abelian
suppose G is an abelian finite group
and let G_p be the p-primary part
that is, the subgroup of elements killed by some power of p
Kannappan Sampath
Kannappan Sampath
Mar 10, 2012 02:39
OK.
Mariano Suárez-Alvarez
Mariano Suárez-Alvarez
then G is the direct sum of the G_p
and Aut(G) is the direct product of the Aut(G_p), because an autom. of G must map each G_p into itself.
Kannappan Sampath
Kannappan Sampath
Needs a proof but I think I am seeing one. So, ok.
Mariano Suárez-Alvarez
Mariano Suárez-Alvarez
So if G is finite abelian with abelian Aut, then each G_p satisfies the same conditions
of course it needs a proof :)
and conversely, if all the G_p satisfy the conditions, the observations made imply that G does too
Kannappan Sampath
Kannappan Sampath
Ah, yes. You put all of this under that carpet? : )
Mariano Suárez-Alvarez
Mariano Suárez-Alvarez
no, I am trying not to ruin all the fun for you :)
 
Conversation ended Mar 10, 2012 at 2:41.