Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This question was initially proposed to me by two friends. Given an integer $n$, how many isomorphism classes are there for semidirect products $\mathbb{Z}/n\mathbb{Z}\rtimes\mathbb{Z}/2\mathbb{Z}$?

Maybe this is a really trivial question. I can tell that a semidirect product is the same as an integer $r\in\mathbb{Z}/n\mathbb{Z}$ with $r^2=1\mod[n]$, but are there isomorphisms between some of them? What happens for instance when n is squarefree, thus the product of fields.

share|improve this question
Such a group is called dihedral only when $r\equiv -1 \mod n$. –  Frieder Ladisch Mar 18 '11 at 11:37
I wasn't aware of that. The question still stands. –  Olivier Bégassat Mar 18 '11 at 11:40
I've corrected the wording. –  Olivier Bégassat Mar 18 '11 at 11:56
add comment

2 Answers

Each Sylow $p$-subgroup of the copy of $\mathbb{Z}/n\mathbb{Z}$ will be normal (actually characteristic) in the semi-direct product. Moreover any element of the semi-direct product not in $\mathbb{Z}/n\mathbb{Z}$ will induce the same automorphism of each Sylow $p$-subgroup since $\mathbb{Z}/n\mathbb{Z}$ is abelian.

Thus two of your semi-direct products will be isomorphic precisely if each of the subproducts $\mathbb{Z}/p^r\mathbb{Z}$ by $\mathbb{Z}/2\mathbb{Z}$ are isomorphic where $p^r$ is the maximal power of $p$ dividing $n$.

In the square-free case this shows that there are $2^k$ non-isomorphic semi-direct products of the form you require, where $k$ denotes the number of odd prime factors of $n$.

In the general case this argument atill reduces you to the case $n$ is a prime power. I guess you'll still get two choices for each odd prime dividing $n$ and two choices for the prime $2$ if $4$ divides $n$ but only one otherwise.

share|improve this answer
I suppose this means that I am claiming that if you have $r\neq r'$ in $\mathbb{Z}/n\mathbb{Z}$ there will be no isomorphisms between the corresponding semi-direct products. You can actually see this directly by observing that you can recover $r$ from the semi-direct product at least if $n$ is odd by realising that any element of order $2$ in the semi-direct product must act as multiplication by $r$ on any normal subgroup of order $n$. –  Simon Wadsley Mar 18 '11 at 12:28
Actually, for the prime 2 you get four choices if 8 divides $n$, two choices if 4 divides $n$ but not 8, and only one choice (or rather: none :) if $n$ is odd. –  Max Horn Mar 18 '11 at 12:28
Right. Thanks. There are 4 roots of $-1$ mod $8$. –  Simon Wadsley Mar 18 '11 at 12:31
There are 4 (square) roots of 1 mod 8. There are no (square) roots of -1 mod 8. –  KConrad Mar 18 '11 at 16:34
Yes... I was clearly not thinking very clearly when I posted this. –  Simon Wadsley Mar 18 '11 at 23:33
add comment

You are asking for isomorphism classes of split extensions $H$ of the module $N:=\mathbb{Z}/n\mathbb{Z}$ by the group $G:=\mathbb{Z}/2\mathbb{Z}$. This is a special case of a metacyclic group, by the way.

A first approximation is to determine all homomorphisms from $\mathbb{Z}/2\mathbb{Z}$ into the automorphism group $Aut(N)$ of $N$; each of these corresponds to a semidirect product (but we still might get some isomorphism classes multiple times). Now, it is well-known that $Aut(N)\cong \mathbb{Z}/\phi(n)\mathbb{Z}$, where $\phi$ denotes Euler's totient function.

Using the above (and the links I gave), it is not difficult to see that the 2-Sylow-subgroup $P$ of $Aut(N)$ is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^m \times Aut(\mathbb{Z}/2^k\mathbb{Z})$, where $m$ equals the number of distinct odd primes divisors of $n$, and $2^k$ is the largest power of $2$ dividing $n$. If $k=0$ or $k=1$, then $P\cong(\mathbb{Z}/2\mathbb{Z})^m$. If $k>1$, then $P\cong (\mathbb{Z}/2\mathbb{Z})^{m+1} \times \mathbb{Z}/2^{k-2}\mathbb{Z}$.

Counting the number of homomorphisms from $G$ into this, then for $k=0$ or $k=1$ we get $2^m$; for $k=2$ we get $2^{m+1}$ and for $k>2$ we get $2^{m+2}$.

With some more effort, one proceeds to verify that each of these homomorphisms leads to a unique isomorphism class; for that, you essentially have to verify that $G$ either acts trivially or non-trivially on each $p$-Sylow-subgroups; and that it really has four non-isomorphic actions on $\mathbb{Z}/2^k\mathbb{Z}$ if $k>2$ (once you get an action like in a dihedral group, once like in a semidihedral / quasidihedral group; once as in the direct product; and one more).

share|improve this answer
Hi Max, I think your $2$ Sylow formula might be wrong. I agree with what you've said before that. Set $n=2^k\cdot\prod_{i=1}^r p_i^{m_i}$ the decomposition into prime factors. Then $\mathrm{Aut}(N)\simeq (\mathbb{Z}/n\mathbb{Z})^{\times}\simeq (\mathbb{Z}/2^k\mathbb{Z})^{\times}\times\Prod_i (\mathbb{Z}/p_i^{m_i}\mathbb{Z})^{\times}$. The factor corresponding to the prime $2$ depends on wether $k\leq 2$, but is a $2$ group. For $p_i$ odd prime however $(\mathbb{Z}/p_i^{m_i}\mathbb{Z})^{\times}\simeq \mathbb{Z}/(p_i-1)\mathbb{Z}\times\mathbb{Z}/p_i^{m_i-1}\mathbb{Z}$. –  Olivier Bégassat Mar 18 '11 at 21:15
so, I think the formula for the $2$ Sylow should be $(\mathbb{Z}/2^k\mathbb{Z})^{\times}\times\prod_{i=1}^r \mathbb{Z}/2^{\nu_i}\mathbb{Z}$ where $\nu_i$ is the highest exponent of $2$ in $P_i-1$. Correct me if I'm wrong, but I think this is the correct formula. I just noticed that I changed your notation, what you call $m$ I call $r$. –  Olivier Bégassat Mar 18 '11 at 21:21
For instance your formula yields that the $2$ Sylow of $\mathbb{Z}/17\mathbb{Z}$ should be $\mathbb{Z}/2\mathbb{Z}$ while it actually is $\mathbb{Z}/16\mathbb{Z}$. –  Olivier Bégassat Mar 18 '11 at 21:24
Olivier, you are absolutely right, my formula for the 2-Sylow as given is incorrect. Luckily, the final result is still correct, as it only depends on the the subgroup of all elements of order 2. This still has the shape $\mathbb{Z}/2\mathbb{Z})^{m+a}$, where $a=0,1,2$ depending on whether $k=0,1$ or $k=2$ or $k>2$. –  Max Horn Mar 21 '11 at 13:03
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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