When are two semidirect products of two cyclic groups isomorphic

Question

(I have posted this question in Math Stack Exchange, only to have received no answer.)

It is well known that a semidirect product of two cyclic groups $C_m$ and $C_n$ has the form $$ C_m \rtimes_k C_n = \langle x,y \mid x^m = y^n = 1,\, yxy^{-1} = x^k \rangle, $$ for some $k^n \equiv 1\pmod m$.

A question that seems quite elementary and yet to which I have not found a satisfactory answer is: what $k$'s give isomorphic semidirect products? Clearly, if $\gcd(r,n)=1$, then $C_m\rtimes_k C_n$ is isomorphic to $C_m\rtimes_{k^r} C_n$ via the mapping $(x,y)\mapsto(x,y^{r^{-1}\ \mathrm{mod}\ n})$. Is the converse also true, that is, if $C_m\rtimes_k C_n$ is isomorphic to $C_m\rtimes_{k'} C_n$, then $k'\equiv k^r\pmod m$ for some $\gcd(r,n)=1$? If so, then we can easily deduce the number of nonisomorphic semidirect products of $C_m$ and $C_n$.

I strongly believe that this is true, but using brute force to verify it seems infeasible. Any help/reference appreciated.

Answer 1

The paper of Basmaji, "On the isomorphisms of two metacyclic groups" (Proc AMS 1969) gives a complete answer to the question of when two finite metacyclic groups with the same $m$ and $n$ are isomorphic, and the paper of Hempel, "Metacyclic groups" (Comm Alg 2000) gives a complete classification of isomorphism types. Your question is about the case of a split metacyclic group, and an affirmative answer can be read off from the main theorems there.

Answer 2

This is too long for a comment and solely deals with the case of coprime $m,n$.

Suppose we have an isomorphism $f:C_m\rtimes_k C_n \to C_m\rtimes_k' C_n$. I would like to name the generators $x,y$ on the left and $X,Y$ on the right.

Then the elements of order dividing $m$ form a subgroup both in the source and in the target and hence $f$ must map $C_m$ to $C_m$. Thus $f(x)=X^r$, where $r$ and $m$ are coprime. The strategy is now to postcompose with automorphisms until $f$ looks like isomorphism you described.

First postcompose with $X\mapsto X^{r'},Y\mapsto Y$, where $r'r=1$ mod $m$. Thus we may assume without loss of generality that $f(x)=X$.

Next note that $f$ induced an isomorphism on the quotient after dividing out $C_m$. Thus $f(y)=X^rY^s$, where $s$ and $n$ are coprime. Postcomposing with $X\mapsto X,Y^s\mapsto X^{-r}Y^s$ (looks weird but $Y^s$ is just another generator of $C_n$), we may additionally assume that $f(y)=Y^s$. Now we can figure out how the conjugation by $Y$ acts and see that we have exactly an isomorphism as you described.

However this argument breaks down as soon as $m$ and $n$ are not coprime. There might be very weird maps appearing like the map that swaps coordinates in $C_n\times C_n$ or worse.