(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.

  • $\begingroup$ Why do you believe that it is true? That is, what is your intuition for why there shouldn't be "accidental" isomorphisms? (My intuition, based on almostt nothing, is that there should be plenty of them.) $\endgroup$
    – LSpice
    Sep 27 at 14:24
  • $\begingroup$ If $m$ and $n$ are coprime, this should be true. Then one can use that the isomorphism must map $C_m$ to $C_m$. $\endgroup$ Sep 27 at 14:49
  • $\begingroup$ @LSpice Its just based on some simple observations. The result is trivially true for $n=2$. For $n=3$, if this is true, it should give us $5$ distinct semidirect products for $m=9p$ or $m=pq$ for distinct primes $p,q\equiv 1\pmod 6$, which is indeed the case for some small $m$. Of course this may be false; it's just my hope that things are simple. :) $\endgroup$ Sep 27 at 16:38
  • $\begingroup$ @HenrikRüping Sorry for not quite following you. What isomorphism did you talk about? $\endgroup$ Sep 27 at 16:40
  • 1
    $\begingroup$ Another easy case is when $m$ is an odd prime power. Indeed, in this case, $C_m^\times$ is cyclic, and hence has at most a unique subgroup of each order. More generally this works if for every prime $p$ dividing $n$, the $p$-Sylow of $C_n^\times$ is cyclic. $\endgroup$
    – YCor
    Sep 27 at 23:36

2 Answers 2


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.

  • 1
    $\begingroup$ That's very impressive. Thanks! $\endgroup$ Sep 28 at 17:35

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.

  • $\begingroup$ Thanks for this! $\endgroup$ Sep 28 at 8:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.