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

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.

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

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