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What is the size of the group under composition of linear functions over $\mathbb{F}_q$ where the functions are of the form $f(x)=ax+b$ where $a\neq 0$ and $a,b\in \mathbb{F}_q$?

So I know that $q=p^n$ for some prime $p$. I also know that the size of a group $G$ can be calculated as $|G| = |G_x||Orb(x)|$. But in this case how would I calculate the size of the stabilizer and orbit of $x$?

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  • $\begingroup$ Wouldn't the only stabilizer of $x$ be the identity polynomial? $\endgroup$ – Mark Dec 12 '16 at 0:17
  • $\begingroup$ @Mark Yeah that does make sense, because in this case it is the only function to take any function to itself. But the orbit of $x$ is the set of elements $y$ such that $xg = y$ for some $g\in G$. Then wouldn't that just be $(p^n-1)p^n$? $\endgroup$ – le maste4r Dec 12 '16 at 0:35
  • $\begingroup$ I think so. This should make sense, as we have that $f(x) = ax+b$ with $a\neq 0$ means there are $p^n-1$ choices for $a$, and $p^n$ choices for $b$. If all of these choices are valid, there should be $(p^n-1)p^n$ total elements in $G$. This is assuming that any linear polynomial is in the orbit of $x$, but this appears to be the case, so we should be fine. $\endgroup$ – Mark Dec 12 '16 at 0:43
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Note that the group you defined is isomorphic to the group of the matrices of the form $$\begin{pmatrix} a & b \\0 & 1 \end{pmatrix}$$ so $(p^n-1)p^n$ is the right answer.

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