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If $G$ is a finite group, a bijection $f\colon G\to G$ is called a (normed) $\boldsymbol{\mathfrak{R}}$-automorphism if

  1. $f$ maps subgroups of $G$ to subgroups of $G$, and

  2. $f(gH) = f(g) f(H)$ for any $g\in G$ and $H \leq G$, i.e. $f$ maps cosets of $H$ to cosets of $f(H)$.

Any automorphism of a group is an $\mathfrak{R}$-automorphism, but the converse does not hold. For example, if $G$ is a cyclic group of prime order, then any permutation of the elements that fixes the identity is an $\mathfrak{R}$-automorphism.

In my answer to this question, I observed that the group $$ \mathbb{Z}_9 \rtimes \mathbb{Z}_3 \;=\; \langle a,b\mid a^9=b^3=1,b^{-1}ab = a^4\rangle $$ has an $\mathfrak{R}$-automorphism defined by $$ f(a^jb^k) = a^{-j}b^{-k} $$ for all $j,k$. My question is

Is there any simple way to prove that $f$ is an $\mathfrak{R}$-automorphism?

The only proof I have is by brute force: we write down all the cosets of all of the subgroups and observe that $f$ maps cosets to cosets.

More generally,

Are there any good techniques for proving that a given function is an $\mathfrak{R}$-automorphism?

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Jim Belk
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  • Do we need to assume $G$ is finite, as is done in the linked question? – Robert Lewis Jul 13 '15 at 00:01
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    @RobertLewis Not per se, although I'm mostly interested in the finite case. In general, the definition of a (normed) $\mathfrak{R}$-automorphism is any automorphism of the coset lattice that fixes the trivial subgroup. This should be the same as the definition I gave in the case of finite groups, but I suppose it might be different for infinite groups. I'll add a finiteness assumption to the question. – Jim Belk Jul 13 '15 at 00:03
  • In the question you linked one has the additionally property that $f$ fixes all subgroups. Does this help in any way? – j.p. Jul 14 '15 at 09:52
  • @j.p. The example above fixes all subgroups as well, but I still don't see any method beyond brute force to check that it has the required property. – Jim Belk Jul 14 '15 at 14:44

1 Answers1

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For the specific example you can do some computations in the group to establish that all subgroups have to satisfy a certain property for the coset condition to be satisfied. This removes the brute force on the left cosets (but doesn't impact the rest).

Let $g=a^j b^k$ and $h=a^l b^m\in H$. Then you have $$f(gh) = f(a^{j+4^kl}b^{k+m}) = a^{-j-4^kl}b^{-k-m}.$$ We want to find $a^x b^y$ such that $f(gh)=a^{-j}b^{-k}a^{-x}b^{-y}$. Computing in $G=\mathbb Z_3\rtimes\mathbb Z_9$ we find that $y\equiv m\bmod 3$ and $x\equiv 4^{2k}l\bmod 9$.

So supposing you have shown that $f$ acts as a permutation on the set of all subgroups, then $f$ will be an $\mathfrak R$-automorphism if and only if for all subgroups $H$ and all $k,l,m\in\mathbb N$ we have $a^l b^m\in H \iff a^{4^{2k}l}b^m\in H$.

For the general case, I do not at this juncture see how to approach the problem generically, and could only use ad hoc computations for specific (types of) groups and choices of $f$.

zibadawa timmy
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