Order of Automorphism Group of $\mathbb{Z}_5\times \mathbb{Z}_5$

Question

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.

See this duplicate. You can also easily find the order of $GL(2,p)$ on this site. — Dietrich Burde, Oct 01 '19 at 15:15
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 $ad-bc\neq 0.$ — Thomas Andrews, Oct 01 '19 at 15:20
In particular, the automorphism group has $24\cdot 20$ elements. — Thomas Andrews, Oct 01 '19 at 15:21
Also, calculating $\phi,$ you have $\phi(25)=20$ since $5,10,15,20,$ and $25$ are not relatively prime to $25.$ — Thomas Andrews, Oct 01 '19 at 15:23
@ThomasAndrews I don't understand how you get $24 \cdot 20$, can you please explain that? — Peter_Pan, Oct 01 '19 at 15:23
@ThomasAndrews, I didn't realize you include $m$ in $\phi(m)$, thanks — Peter_Pan, Oct 01 '19 at 15:24

score 2 · Accepted Answer · answered Oct 01 '19 at 16:55

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 $GL_2(\mathbb{Z}_5)$.

Order of Automorphism Group of $\mathbb{Z}_5\times \mathbb{Z}_5$

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