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Part of the answer to this question: Group of order $1575$ problem

involves the fact that $\mid Aut(Z_3 \times Z_3) \mid=48$. I only know of a corollary dealing with the order of the automorphism group of cyclic groups, which $Z_3 \times Z_3$ is not. Am I missing something obvious here or is this not trivial to know?

Community
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Mike
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  • This paper seems useful for your question (although may not provide a specific solution. I've only read the first page so far, and it's only dealing with relatively prime order $G\times H$). – Mark Schultz-Wu Dec 14 '16 at 01:58
  • Interesting, I will definitely come back to read this, but their result is definitely not something a student of algebra could carry around and bust out at a moment's notice(!), as the people do in the linked page.

    This is a small part of a small question out of tens of questions in section 4.4 on Automorphisms in Dummit & Foote. I'm thinking can't possibly require such deep thought, which leads me to believe I'm missing something very simple here.

     – Mike Dec 14 '16 at 02:02

1 Answers1

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Well, they are the same as the number of vector space automorphisms from $\mathbb F_3^2$ to $\mathbb F_3^2$. Which is the same as the number of invertible $2\times 2$ matrices in $\mathbb F_3$.

There are $8$ options for the first column of such a matrix, and then the second column must not be a multiple of the first column, so there are $9-3=6$ options. So there are $8\times 6=48$ automorfisms.

Asinomás
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    You can strip away the vector space stuff and just say that you can map $(1,0)$ and $(0,1)$ independently and this defines uniquely every morfism. But you need $f(0,1)$ and $f(1,0)$ to generate the whole group. So $f(1,0)$ cannot be zero, and $f(0,1)$ cannot be in the subgroup generated by $f(1,0)$. So there are $8\times 6$ options. – Asinomás Dec 14 '16 at 02:04
  • Thank you for your answer. I imagine $F^2_3$ is notation for the vector space of 2-tuples over $Z/3Z$, correct?

    Then wouldn't the choices for the first column be $$(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)$$

    EDIT: Duh, $(0,0)$ can't be there.

    Great, thanks!

     – Mike Dec 14 '16 at 02:10
  • yes, the notation is correct, but notice that $(0,0)$ is not an option, because we want the column space to have dimension $2$ (remember that a square matrix of size $n$ is invertible if and only if the column space has dimension $n$) – Asinomás Dec 14 '16 at 02:12
  • Very Elegant Answer Sir But I have doubt .I understand that 48 automorphism of $Z_3\times Z_3$ which equals order of that Matrix.But How to show they are Isomorphic to Each other ?@Jorge Fernández – Curious student May 03 '18 at 06:23