I am trying to characterize when a semi-direct product of the form $(Z/pZ)^n \rtimes (Z/qZ)$ is isomorphic to a group generated by two elements. Here $p$ and $q$ are distinct odd primes.

I would be happy for a reference or even some examples of this happening for $n > 1.$

(I found a similar question here in the case that $p=2$, but the odd case may be considerably different, and I am looking for different structural information any way.)


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    $\begingroup$ One example where this happens is if the corresponding representation of $\mathbb{Z}/q\mathbb{Z}$ is irreducible, which can happen precisely when $n$ is the multiplicative order of $p$ mod $q$. $\endgroup$ – Tobias Kildetoft Aug 15 '18 at 7:16
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    $\begingroup$ Write $V=(Z/pZ)^n$ as direct sum of irreducibles $V_i^{n_i}$ over $Z/qZ$, where the $V_i$ are pairwise non-isomorphic as $Z/qZ$-modules. Write $V_1=Z/pZ$ viewed as trivial $Z/qZ$-module. Write $m_i=n_i$ for $i>1$ and $m_i=n_i-1$. Then your group $G$ is generated by two elements if and only if $m_i\le 1$ for all $i$ (that is, $n_1\le 2$ and $n_i\le 1$ for $i>1$). Here $p$ is any prime and $q$ can be any power of a prime $\neq p$. I'll write details later but it's a good exercise. $\endgroup$ – YCor Aug 15 '18 at 7:19
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    $\begingroup$ @PatDevlin You have some choice in the orders of the two generators. For example, if there are no trivial modules ($n_1=0$) then there is a choice between two generators of order $p$, or one of order $p$ and the other of order $q$. If $n_1=2$, then you must choose one generator of order $pq$ and the other can be of order $pq$ or $q$. There is even more choice when $n_1=1$. $\endgroup$ – Derek Holt Aug 15 '18 at 16:56
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    $\begingroup$ @LucGuyot Derek already answered: according to whether $n_1\le 1$ or $n_1=2$, write $G=V\rtimes Z/qZ$ or $G=V\rtimes Z/pqZ$ with $V$ cyclic (i.e. with all multiplicities $\le 1$). Then use one generator of the right-hand cyclic group and one of the cyclic module. $\endgroup$ – YCor Aug 16 '18 at 5:13
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    $\begingroup$ It may be of some interest to observe that the group is 2-generated if and only if it is a quotient of $C_q \times (C_q \wr C_p)$. $\endgroup$ – Derek Holt Aug 16 '18 at 7:02

This answer corroborates YCor's claim according to which the conditions $n_i \le 1$ for $i > 1$ and $n_1 \le 2$ on the irreducible modular representations'multiplicities $n_i$, are necessary and sufficient for $G$ to be two-generated. We actually show a slightly more general result expressed in terms of the geometric multiplicity of the eigenvalue $1$.

Let $K = Z/pZ, V = K^n$ and $C = Z/qZ$ with $p$ a prime number and with $q \ge 2$ an integer. We fix a group homomorphism $\varphi: C \rightarrow GL(V)$ and a favored generator $a$ of $C$. Let $G = V \rtimes_{\varphi} C$ the corresponding semi-direct product.

The $K$-vector space $V$ is endowed with the structure of $K[X]$-module induced by $X \cdot v = \varphi(a)(v)$. If $V$ is a cyclic $K[X]$-module generated by $w$, then $G$ is generated by $(w, a)$, hence two-generated. The converse holds if $G$ is centerless.

Claim 1. Assume that the center of $G$ intersects $V$ trivially, i.e., $\varphi(a)$ has no non-zero fixed vector. If $G$ is two-generated then $V$ is a cyclic $K[X]$-module.

Proof. Since $\varphi(a)$ has no non-zero eigenvector associated to $1$, the minimal polynomial $\mu(X)$ of $\varphi(a)$ over $K$ is coprime with $X - 1$. As $\mu(X)$ divides $X^q - 1$, this minimal polynomial divides $\nu(X) = 1 + X + \cdots + X^{q -1}$. Let $(v_1a^{n_1}, v_2a^{n_2})$ be a generating pair of $G$ with $v_i \in V, n_i \in \mathbb{Z}$ for $i = 1,2$. We can reduce this pair to a generating pair of the form $(v, wa)$ with $v,w \in V$ by means of Nielsen transformations (consider the induced transformations on $C^2$). Given $u \in V$, we shall show that $u \in K[X] \cdot v$. Since $(v, wa)$ generates $G$, there is a word $x$ on the alphabet $\{x_1^{\pm 1}, x_2^{\pm 1}\}$ such that $u = x(v, wa)$. Write $x = yx_2^{s}$ where $y$ belongs to the normal closure of $x_1$ in the free group $F(x_1, x_2)$ and $s$ lies in $\mathbb{Z}$. Note that we have $y(v, wa) \in K[X]\cdot v$ because $K[X]\cdot v$ is normalized by $wa$. It follows that $s = kq$ for some $k \in \mathbb{Z}$ and hence $u = y(v, wa) + k \nu(X) \cdot w = y(v, wa)$, which completes the proof.

Checking whether $V$ is a cyclic $K[X]$-module can be done algorithmically by computing the Smith Normal Form of $\varphi(a)$ [Theorem 20 of Section 12, 1].

If the center $Z(G)$ of $G$ has a non-trivial intersection with $V$, then $G/(Z(G) \cap V)$ yields a semi-direct product decomposition $V' \rtimes C$ where $V'$ is a finite-dimensional vector space over $K$ of lower dimension. The space $V'$ is obtained as the direct sum of the non-trivial irreducible components of the representation $\varphi$ when $p$ does not divide $q$ as modular representations are completely reducible in this case. If $G$ is moreover two-generated, then $V'$ must be a cyclic $K[X]$-module. In addition, $Z(G)$ must be two-generated.

I am indebted to Derek Holt and Geoff Robinson for the latter remark which is expanded below.

If $G$ is the semi-direct product $V \rtimes C$ of an arbitrary Abelian group $V$ with an arbitrary cyclic group $C$, then it is easily checked that $Z(G) = (Z(G) \cap V) \times (Z(G) \cap C)$. In the context of the question, $C$ is simple so that $Z(G) \cap C$ is either equal to $C$, in which case $G$ is Abelian, or trivial. If moreover $p$ does not divide $q$, it follows that $Z(G)$ is a direct factor of $G$. Thus, if $G$ is two-generared, so is $Z(G)$.

The above claim and the subsequent remark allows us to recover

YCor's claim. Assume that $p$ does not divide $q$. Then the following are equivalent:

  • $G$ is two-generated;
  • the multiplicity of any non-trivial irreducible representation in $\varphi$ is at most $1$, the multiplicity of the trivial representation is at most $2$.

Proof. In order to show that the condition is sufficient, we can reduce to the case of a cyclic $K[X]$-module by moving out of $V$ one $Z/pZ$-factor of $V$ with trivial $Z/qZ$-action if necessary, noting that $Z/pZ \times Z/qZ \simeq Z/pqZ$. The fact that the condition is necessary follows from Claim 1.


The following generalization relies on the fact that the restriction of $\varphi(a)$ to the generalized eigenspace associated to $1$ has a Jordan normal form.

Claim 2. Let $p$ be a prime number and let $q \ge 2$ be an integer. Let $A = \varphi(a)$ and let $\mu_A$ be the minimal polynomial of $A$ over $K$. Write $\mu_A(X) = (X - 1)^k P(X)$ with $P(X) \in K[X]$ and $P(1) \neq 0$. Let $I \in GL_n(K)$ be the identity matrix and let $E = \ker((A - I)^k)$. Let $\gamma_1(A) = \dim_K(\ker(A - I))$. We set $$ d(A) = \left\{ \begin{array}{cc} \gamma_1(A) + 1& \text{ if } p \text{ divides } q, \\ \gamma_1(A) & \text{ otherwise. } \end{array} \right. $$ Then the group $G = V \rtimes_{\varphi} C$ is two-generated if and only if $V/E$ is a cyclic $K[X]$-module and $d(A) \le 2$.

Proof. The abelianization of $G$ surjects onto an elementary Abelian $p$-group of rank $d(A)$. Thus the condition on $d(A)$ is necessary. By Claim 1, the quotient $G/E$ is two-generated if and only if $V/E$ is a cyclic $K[X]$-module. Now, we only need to verify that $G$ is two-generated when $d(A) = 2$ and $p$ does not divide $q$. Indeed, for the other cases, the restriction of $\varphi(a)$ to $E$ has at most one Jordan block so that $V$ is a cyclic $K[X]$-module if $V/E$ is. If $d(A) = 2$ and $p$ doesn't divide $q$, then the restriction of $\varphi(a)$ to $E$ must be the identity, since otherwise its multiplicative order, which is a divisor of $q$, would be a multiple of $p$ (consider the top left $2$-by-$2$ sub-matrix of a Jordan block). We conclude as before, by moving out of $V$ one of the two point-wise $C$-invariant $Z/pZ$-factors.

[1] D. Dummit and R. Foote, "Abstract Algebra", 1999.

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    $\begingroup$ I don't believe that your final sentence is correct. $Z(G)$ can have order $q^2$ and you can choose generators of order $pq$. $\endgroup$ – Derek Holt Aug 15 '18 at 18:04
  • $\begingroup$ But $X$ is acting trivially on $Z(G)$ so that makes no difference. $\endgroup$ – Derek Holt Aug 15 '18 at 18:25
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    $\begingroup$ You might remark that in the situation of the question, the center is a direct factor of the group. $\endgroup$ – Geoff Robinson Aug 15 '18 at 21:20
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    $\begingroup$ Leaving part of the proof to the comments is not recommended, as comments can be deleted at any time, and it makes it painful for readers to find what they want even if they remain (especially readers that weren't actively monitoring the Q&A as it unfolded). $\endgroup$ – zibadawa timmy Aug 17 '18 at 6:38
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    $\begingroup$ @zibadawatimmy I agree with you. I have made changes accordingly. $\endgroup$ – Luc Guyot Aug 17 '18 at 6:44

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