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Let $R$ be the ring of $3 \times 3$ matrices with coefficients in $\Bbb Z_5$.
For every $g \in GL_3(\Bbb Z_5)$ prove that the function $$f\colon R \rightarrow R$$ defined as $$x \mapsto g^{-1}xg$$ is an automorphism of $R$.

If I choose the matrix $$g = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix},$$ how many matrices $x\in R$ such that $f(x)=x$ are there?

I have no idea how to solve this. Any help?

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    $\begingroup$ Verify that $f(xy) = f(x)f(y)$ and $f(x + y) = f(x) + f(y)$ for all matrices $x$ and $y$ of $R$. Show that $f$ is bijective. $\endgroup$ – M. Vinay Mar 20 at 17:44
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    $\begingroup$ After that, determine $g^{-1}$ for that particular matrix $g$ (hint: compute $g^2$). Now, describe in words what left-multiplication by $g^{-1}$ does to a $3 \times 3$ matrix (note that $g$ is a permutation matrix). Describe what right-multiplication by $g$ does. Combine these two to describe the relation between $f(x) = gxg^{-1}$ and $x$ for any matrix $x$. Is it clear now what the structure of $x$ must be in order that $f(x) = x$? $\endgroup$ – M. Vinay Mar 20 at 17:48
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    $\begingroup$ I think that I should consider x and y in R. Only g should be considered in $GL_3(\Bbb Z_5)$ $\endgroup$ – Jack Mar 20 at 17:48
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    $\begingroup$ Yes, comment fixed. $\endgroup$ – M. Vinay Mar 20 at 17:49
  • $\begingroup$ Just to be sure in an effort to do what I'm doing, I'm new to it. Let's consider $f(x + y) = f(x) + f(y)$. $f(x+y)= g^{-1}(x+y)g$. Is that correct? So $f(x)+f(y)= g^{-1}xg+g^{-1}yg$ and they are equal since $f(x+y)=g^{-1}xg+g^{-1}yg$ by doing multiplications. Same approach for $f(xy) = f(x)f(y)$. Is that correct? $\endgroup$ – Jack Mar 20 at 18:00
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This is much more general: suppose $R$ is a ring and $g$ is invertible in $R$. Then the map $f\colon R\to R$, $f(x)=gxg^{-1}$ is an automorphism of $R$.

Indeed, $f(x+y)=g(x+y)g^{-1}=gxg^{-1}+gyg^{-1}=f(x)+f(y)$ and $$ f(xy)=gxyg^{-1}=gxg^{-1}gyg^{-1}=f(x)f(y) $$ Obviously, $f(1)=1$.

Since the map $x\mapsto g^{-1}xg$ is the inverse of $f$, we are done.

For the second part, you want $gxg^{-1}=x$, so $gx=xg$. If $$ x=\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} $$ then $$ gx=\begin{pmatrix} a_{31} & a_{32} & a_{33} \\ a_{21} & a_{22} & a_{23} \\ a_{11} & a_{12} & a_{13} \\ \end{pmatrix} \qquad xg=\begin{pmatrix} a_{13} & a_{12} & a_{11} \\ a_{23} & a_{22} & a_{21} \\ a_{33} & a_{32} & a_{31} \end{pmatrix} $$ Thus you get \begin{cases} a_{31}=a_{13} \\ a_{32}=a_{12} \\ a_{33}=a_{11} \\ a_{21}=a_{23} \end{cases} and so the linear system has rank $4$.

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  • $\begingroup$ Thanks, for your answer. It helped me a lot. But I would like to clarify some steps. What is the reasoning behind $x\mapsto g^{-1}xg$ is the inverse of $f$? Then, why you put $gx=xg$? And finally why the linear system and how can it help to find how many matrices satisfies the request? I hope I have expressed my doubts correctly. Thanks. $\endgroup$ – Jack Mar 21 at 11:05
  • $\begingroup$ @Jack Conjugating by $g^{-1}$ undoes what conjugating by $g$ does. For the rest: $gxg^{-1}=x$ is equivalent to $gx=xg$. The matrix $g$ swaps rows 1 and 3 if at the left, columns 1 and 3 if at the right. The linear system has rank $4$, so the null space has dimension $9-4=5$. How many vectors are there in a vector space of dimension $5$ over a field with five elements? $\endgroup$ – egreg Mar 21 at 11:50
  • $\begingroup$ I'm sorry to ask, but would there be an alternate way to solve the last part? Vectors and vector spaces are subjects I don't know about. Maybe that's why I couldn't understand your resolution. Since this exercise refers to a part of the theory where vector spaces are not mentioned I think there is an alternative way to solve it. Thank you for your understanding. $\endgroup$ – Jack Mar 21 at 13:04
  • $\begingroup$ @Jack You’re free to choose five entries, namely those listed in the left hand sides plus $a_{22}$. How many choices can you make? $\endgroup$ – egreg Mar 21 at 16:18
  • $\begingroup$ $5^5$ I suppose since we are working on $\Bbb Z_5$ (and if I correctly understood what you mean) $\endgroup$ – Jack Mar 21 at 16:31

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