From: mareg@primrose.csv.warwick.ac.uk ()
Subject: Re: Infinite cyclic automorphism group
Date: Sun, 2 Dec 2001 12:39:15 +0000 (UTC)
Newsgroups: sci.math
Summary: Groups with free-abelian automorphism group

In article <73p1h8zfm9rk@legacy>,
	bobkatz@hotmail.com (Bob Katz) writes:
>> Is there a group G such that Aut(G) is infinite cyclic group ?
>
>>No. Such a group would have G/Z(G) cyclic and this implies
>>that G is abelian. Every abelian group apart from the 2-torsion
>>groups has an automorphism of order 2 and 2-torsion abelian
>>groups of order >= 4 also have automorphisms of order 2.
>
>>Robin Chapman
>
>OK thanks, but let me ask also if Aut(G) can be ZxZ (direct product of
>two infinite cyclic groups) .
>

No. All subgroups of a free abelian group are free abelian, so a subgroup
of ZxZ is isomorphic  to 1, Z or ZxZ. Hence G/Z(G) is is isomorphic to
one of these. Robin has proved that 1 and Z are impossible,
so this only leaves G/Z(G) =~ ZxZ. Let a and b in G map onto the
two generators of ZxZ, and let c = [b,a] = b^-1 a^-1 b a be the commutator.

Every element of the group can be uniquely represented as  a^i b^j x, for
x in Z(G).  Now, since commutators are central, the commutator map is
bilinear - i.e.  [b^i,a^j] = c^{ij}  for all  i,j in Z. Hence multiplication
in G is given by the rule
(a^i b^j x)(a^k b^l y) = a^{i+k} b^{j+l} c^{jk} xy, and it is now  easy
to check that G has an automorphism of order 2 given by
a^i b^j x -> a^{-i} b^{-j} x.

I think the argument extends in then obvious way to show that Z^n cannot
occur as Aut(G) for any n.

Derek Holt.