Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
I'm actually more interested in the special case which I wrote above as the "context" for the first question. –  oxeimon Jan 20 '13 at 16:14

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer
    
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). –  oxeimon Jan 20 '13 at 18:03
    
thanks! I can't believe I didn't think to check that. My intuition for semidirect products is terrible. –  oxeimon Jan 20 '13 at 18:04
    
@oxeimon you're welcome! –  Andreas Caranti Jan 20 '13 at 20:41

Your Answer

 
discard

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.