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I want to find the order of $\operatorname{Aut}(\mathbb{Z}_5\times\mathbb{Z}_5)$.

Since generators need to map to generators, and the identity needs to map to the identity, and since any non-identity element generates this group, I think the order should be $4 \cdot3 \cdot 2 = 24$.

But I saw here that

$\left| \operatorname{Aut}(G)\right|=\phi(m)$ where $\phi(m)$ is Euler's function, and $m$ is the order of the cyclic group.

I think the order of $\mathbb{Z}_5\times\mathbb{Z}_5$ is $25$, and $\phi(25)=20$ since 5,10, 15, 20, and 25 are not relatively prime to $25$.

Where am I going wrong?


This is similar to This question, but I don't understand the answer.

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    $\begingroup$ $\mathbb{Z}_5 \times \mathbb{Z}_5$ is not cyclic. $\endgroup$
    – lhf
    Commented Oct 1, 2019 at 15:13
  • $\begingroup$ oh. So $|\text{Aut}( \mathbb{Z}_5) | = \phi(5) = 4$ $\endgroup$
    – Peter_Pan
    Commented Oct 1, 2019 at 15:14
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    $\begingroup$ See this duplicate. You can also easily find the order of $GL(2,p)$ on this site. $\endgroup$ Commented Oct 1, 2019 at 15:15
  • $\begingroup$ Nope, you can get $\phi(1,0)=(a,b)$ and $\phi(0,1)=(c,d)$ for any $a,b,c,d\in\mathbb Z_5$ where $ad-bc\neq 0.$ $\endgroup$ Commented Oct 1, 2019 at 15:20
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    $\begingroup$ Also, calculating $\phi,$ you have $\phi(25)=20$ since $5,10,15,20,$ and $25$ are not relatively prime to $25.$ $\endgroup$ Commented Oct 1, 2019 at 15:23

1 Answer 1

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The group $\mathbb{Z}_5$ is a field, $V=\mathbb{Z}_5\times\mathbb{Z}_5$ is a 2-dimensional vector space over $\mathbb{Z}_5$, and any automorphism of $V$ (as a group) is automatically linear. Therefore, $|\mathrm{Aut}(V)|=|GL_2(\mathbb{Z}_5)|$.

To find $|GL_2(\mathbb{Z}_5)|$, we'll do a count. Note that the columns of any $A\in GL_2(\mathbb{Z}_5)$ form a basis for $V$. To count the number of such $A$, note that the first column of $A$ can be any nonzero vector (there are $24$ of those). Given the first column of $A$, the second column can be any vector not in the span of the first column (there are $25-5=20$ of those). Therefore, there are $24\cdot 20=480$ matrices in $GL_2(\mathbb{Z}_5)$.

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