# How tall is the automorphism tower of a group?

[bibtex key=Hamkins2001:HowTall?]

The automorphism tower of a group is obtained by computing its automorphism group, the automorphism group of that group, and so on, iterating transfinitely by taking the natural direct limit at limit stages. The question, known as the automorphism tower problem, is whether the tower ever terminates, whether there is eventually a fixed point, a group that is isomorphic to its automorphism group by the natural map. Wielandt (1939) proved the classical result that the automorphism tower of any finite centerless group terminates in finitely many steps. This was generalized to successively larger collections of groups until Thomas (1985) proved that every centerless group has a terminating automorphism tower. Here, it is proved that every group has a terminating automorphism tower. After this, an overview is given of the author’s (1997) result with Thomas revealing the set-theoretic essence of the automorphism tower of a group: the very same group can have wildly different towers in different models of set theory.

## One thought on “How tall is the automorphism tower of a group?”

1. Now that the automorphism tower problem is solved, everyone can now work on the other tower problem, namely the congruence tower problem (the congruence tower is quite central to point-free topology ). The lattice of all congruences of a frame https://en.wikipedia.org/wiki/Pointless_topology is always a frame, and one can repeat this process transfinitely. Sometimes this transfinite process never ends, but in all known cases it either never ends or ends before the fourth step (and when the process terminates, it always terminates at a complete Boolean algebra).

The cool thing about the congruence tower problem is that it is downwards absolute under forcing; if there is a frame L in a forcing extension V[G] whose congruence tower terminates at an ordinal alpha, then there is a frame in M in V whose congruence tower also terminates at the same ordinal alpha.