Suppose I have two groups $G,H$ and a homomorphism $H\rightarrow\text{Aut}(N)$. Suppose further that I have another group $T$ and two maps: $f_G : G\rightarrow\text{Aut}(T)$ and $f_H : H\rightarrow\text{Aut}(T)$.

Does this induce a map $f : G\rtimes H\rightarrow\text{Aut}(T)$ that restricts to $f_G,f_H$ on $G$ and $H$?

The context for this question is as follows. Let $N$ be an ideal of $\mathbb{Z}[\zeta_d]$. Then the multiplicative group $(\mathbb{Z}[\zeta_d]/N)^*$ acts on its additive group $\mathbb{Z}[\zeta_d]/N$ by multiplication. Thus, we may form the semidirect product $\mathbb{Z}[\zeta_d]/N\rtimes (\mathbb{Z}[\zeta_d]/N)^*$. Clearly the first factor acts on $\mathbb{Z}[\zeta_d]/N$ by translation, and the second factor acts on $\mathbb{Z}[\zeta_d]/N$ as mentioned before by multiplication, thus yielding maps $$\mathbb{Z}[\zeta_d]/N\rightarrow\text{Aut}(\mathbb{Z}[\zeta_d]/N),\qquad\text{and}\qquad (\mathbb{Z}[\zeta_d]/N)^*\rightarrow\text{Aut}(\mathbb{Z}[\zeta_d]/N).$$ (here by Aut I mean automorphism groups of the set)

Gabriel Berger in section 2.3 of his paper "Fake Congruence Modular Curves and Subgroups of the Modular Group" says that this induces an action of $\mathbb{Z}[\zeta_d]/N\rtimes (\mathbb{Z}[\zeta_d]/N)^*$ on $\mathbb{Z}[\zeta_d]/N$ (as a set). However, I don't see how this action can work (his description of the action doesn't seem to be a legit action).

In other words, our $G$ is the additive group of $\mathbb{Z}[\zeta_d]/N$, and $H$ is the multiplicative group of that ring. Then multiplication in the semidirect product is done as follows: $(g_1,h_1)(g_2,h_2) = (g_1 + h_1g_2, h_1h_2)$ (actually his multiplication I think is the opposite of this, but neither multiplication works to make the action that he describes a legit group action).

Now take $t\in\mathbb{Z}[\zeta_d]/N$, and consider the action that he describes (taken as a right-action): $t.(g,h) = (t+g)h$.

Now consider: $t.(g_1,h_1)(g_2,h_2)$.

First, $t.((g_1,h_1)(g_2,h_2)) = t.(g_1+h_1g_2, h_1h_2) = (t+g_1+h_1g_2)h_1h_2 = th_1h_2 + g_1h_1h_2 + h_1^2h_2g_2$

On the other hand, $(t.(g_1,h_1))(g_2,h_2) = ((t+g_1)h_1 + g_2)h_2 = th_1h_2 + g_1h_1h_2 + g_2h_2$

Which are clearly not the same. Pretending that the action is a left-action still doesn't fix things.

  • $\begingroup$ I'm actually more interested in the special case which I wrote above as the "context" for the first question. $\endgroup$
    – oxeimon
    Commented Jan 20, 2013 at 16:14

1 Answer 1


I think all Berger claims is that the semidirect product $\mathbf{Z}[\zeta_{d}] \rtimes \mathbf{Z}[\zeta_{d}]^{*}$ acts on the set $\mathbf{Z}[\zeta_{d}]$. The element $(a,b) \in \mathbf{Z}[\zeta_{d}] \rtimes \mathbf{Z}[\zeta_{d}]^{*}$ acts on $x \in \mathbf{Z}[\zeta_{d}]$ by $x \mapsto b x + a$. It seems to me that this is quite clearly spelled out by Berger, which works in a slightly more general setting.

  • $\begingroup$ Ahhhh, see what he wrote was that $t(x[\gamma]) = (t+x)[\gamma]$ (ie, first add, then multiply), whereas the action you gave was $t(x[\gamma]) = (t\gamma + x)$ (first multiply, then add). $\endgroup$
    – oxeimon
    Commented Jan 20, 2013 at 18:03
  • $\begingroup$ thanks! I can't believe I didn't think to check that. My intuition for semidirect products is terrible. $\endgroup$
    – oxeimon
    Commented Jan 20, 2013 at 18:04
  • $\begingroup$ @oxeimon you're welcome! $\endgroup$ Commented Jan 20, 2013 at 20:41

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