Infinite cyclic automorphism group
Bob Katz
Thanks
Robin Chapman
> 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
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Bob Katz
>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) .
Bob
ma...@primrose.csv.warwick.ac.uk
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.