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Let $G=A\times A$; $A$ be cyclic group of order $p$ where $p$ is a prime .How many automorphisms does $G$ have?

My thoughts:

If we have a cyclic group $G$ of order $n$ then I know that in order for $\phi:G\to G$ be an auto we have that $\phi(x)=x^i;\gcd(i,n)=1$ are the only ones . But I could not follow for this group .Any help

Learnmore
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3 Answers3

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I saw this question late but I feel my contribution is not late.

Note that for sure the identity is mapped to itself. Let $A_1=<(a,e)>$ and $A_2=<(e,a)>$. Observe that if the image of $(a,e)$ is fixed then automatically all other non-identity elements in $A_1$ (i.e.,$(a,e)^i$ with $2\leq i \leq p-1$) have their images determined. $(a,e)$ has $p^2-1$ choices as each non-identity element has order $p$. So all elements in $A_1$ have their images determined. Similarly, we note that $(e,a)$ has $p^2-p$ choices left and all other non-identity elements in $A_2$ have their images automatically fixed. Also note that by fixing images of $(a,e)$ and $(e,a)$ we have automatically fixed the image of an element of the form $(a_1,a_2)$, where $a_1,a_2\neq e$, as it is equal to $(a_1,e)(e,a_2)$. This means that $o(Aut(G))=(p^2-1)(p^2-p)$.

Hope this helps!

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Following up Oliver Braun's answer, the number of automorphisms $G$ has is the order of the group $\text{GL}_2(\mathbb{F}_p)$, which is $(p^2-1)(p^2-p)$.

Edit: Any $n\times n$ invertible matrix over $\mathbb{F}_p$ consists of $n$ linearly independent column vectors. We may construct such a matrix by choosing suitable column vectors one by one. In the $n$-dimensional vector space $\mathbb{F}_p^n$, there are $p^n-1$ nonzero vectors to choose from to be the first column vector. Assume we have chosen $i$ linearly independent column vectors, which span an $i$-dimensional subspace consisting of $p^i$ vectors. Then there are $p^n-p^i$ vectors to choose from to be the $i+1$-st column vector which is linearly independent to the first $i$ chosen column vectors. In sum there are $(p^n-1)(p^n-p)\cdots(p^n-p^{n-1})$ ways to choose $n$ linearly independent column vectors. This number is also the order of $\text{GL}_n(\mathbb{F}_p)$.

Alex Fok
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Identify your group $A\times A$ with the two-dimensional vector space over $\mathbb{F}_p$. Any vector space automorphism of that space is also a group automorphism. Try to work out that the converse also holds. Therefore, $\mathrm{Aut}(A\times A)\cong \mathrm{GL}_2 (\mathbb{F}_p)$.

Notice that this holds more generally, $\mathrm{Aut}(\underbrace{A\times A \times ... \times A}_{n\text{ times}}) \cong \mathrm{GL}_n(\mathbb{F}_p)$.

Oliver Braun
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  • The above argument shows that the automorphism group is isomorphic to $\mathrm{GL}_2(\mathbb{F}_p)$, so their orders coincide. – Oliver Braun May 17 '15 at 06:53
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    Don't give up so fast, you really should spend some more time thinking about this yourself - that's the best way to learn and understand. – Oliver Braun May 17 '15 at 07:04
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    No, I can't chat with you, I really meant it when I said you should spend more time on the problem - maybe a day, so you might ask for more help tomorrow. But realize that this is really basic stuff, something you learn within your first two years of studying mathematics. It has nothing to do with being or becoming a PhD student. Either way, the best (and possibly only) way to make progress on this level is to work out the exercises by yourself and be willing to spend time on them - certainly more than an hour. – Oliver Braun May 17 '15 at 07:12
  • thought over it and came over with this::::Any isomorphism(linear transform) between two vector spaces can be represented by an invertible matrix ;so its same to find number of non-singular matrices – Learnmore May 17 '15 at 14:27
  • That is how you determine the order of $\mathrm{GL}_2(\mathbb{F}_p)$, so this line of thought gives you the number of vector space automorphisms of $\mathbb{F}_p^2$. However, that is something you should already know and it is also somewhat beside the point of this exercise: you need to understand what those vector space automorphisms have to do with the group automorphisms of $A\times A$. By the way, my answer contains everything you need in order to make that connection. – Oliver Braun May 17 '15 at 15:32