Finding $|Aut(\mathbb{Z}_6 \times \mathbb{Z}_{10} \times \mathbb{Z}_{15})|$

Question

I am trying to compute:

$|Aut(\mathbb{Z}_6 \times \mathbb{Z}_{10} \times \mathbb{Z}_{15})|$

I know that:

1. If $H$ and $K$ have coprime orders then:

   $Aut(K \times H) \cong Aut(K)\times Aut(H) $

2. For prime $p$:

   $ GL(\mathbb{Z}_p \times \mathbb{Z}_p) = Aut(\mathbb{Z}_p \times \mathbb{Z}_p) $

I thought of doing something like:

$|Aut(\mathbb{Z}_6 \times \mathbb{Z}_{10} \times \mathbb{Z}_{15})|=|Aut(\mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_{5}\times \mathbb{Z}_{2}\times \mathbb{Z}_{5}\times \mathbb{Z}_{3})|=|Aut(\mathbb{Z}_2^2 \times \mathbb{Z}_3^2 \times \mathbb{Z}_{5}^2)|$

(but why can I swap the order in the last transition?) (Q1)

and then:

$=|Aut(\mathbb{Z}_2^2)|\cdot |Aut(\mathbb{Z}_3^2)| \cdot |Aut(\mathbb{Z}_{5}^2)| $

(but why does this hold? Fact no. 1 is only for two groups and of coprime order...) (Q2)

and from here I want to use fact no. 2 somehow, but I'm not sure how to compute $ |GL(\mathbb{Z}_p \times \mathbb{Z}_p)| $ (Q3)

Any assistance with my questions would be greatly appreciated.

Answer 1

• By the Chinese remainder theorem: $$\mathbb{Z}_6 \times \mathbb{Z}_{10} \times \mathbb{Z}_{15}\\\cong (\mathbb{Z}_2 \times \mathbb{Z}_3) \times (\mathbb{Z}_2 \times \mathbb{Z}_5) \times (\mathbb{Z}_3 \times \mathbb{Z}_5)\\ \cong (\mathbb{Z}_2 \times \mathbb{Z}_2) \times (\mathbb{Z}_3 \times \mathbb{Z}_3) \times (\mathbb{Z}_5 \times \mathbb{Z}_5)$$ On (Q1): We used the pretty obvious commutativity law $G \times H \cong H\times G$ in this step. To work out the details, show that $\phi : G\times H \to H\times G, (g,h) \mapsto (h,g)$ is an isomorphism of groups. Also, we used associativity of $\times$ (up to isomorphism), which is $(G \times H) \times K \cong G\times (H\times K)$, which has a similarly trivial proof.

• As you said, you can split Aut over groups of coprime order, so: $$\operatorname{Aut}(\mathbb{Z}_6 \times \mathbb{Z}_{10} \times \mathbb{Z}_{15}) \\ \cong \operatorname{Aut}((\mathbb{Z}_2 \times \mathbb{Z}_2) \times (\mathbb{Z}_3 \times \mathbb{Z}_3) \times (\mathbb{Z}_5 \times \mathbb{Z}_5)) \\ \cong \operatorname{Aut}(\mathbb{Z}_2 \times \mathbb{Z}_2) \times \operatorname{Aut}(\mathbb{Z}_3 \times \mathbb{Z}_3) \times \operatorname{Aut}(\mathbb{Z}_5 \times \mathbb{Z}_5) \\ \cong \operatorname{GL}(\mathbb{Z}_2 \times \mathbb{Z}_2) \times \operatorname{GL}(\mathbb{Z}_3 \times \mathbb{Z}_3) \times \operatorname{GL}(\mathbb{Z}_5 \times \mathbb{Z}_5) \\ = \operatorname{GL}(2,2) \times \operatorname{GL}(2,3) \times \operatorname{GL}(2,5)$$ Note that the splitting has been done alredy in this step, answering (Q2).

• (Q3) There is the formula $\#\operatorname{GL}(n,q) = \prod_{i=0}^{n-1}(q^n - q^i)$ from linear algebra (counting the number of ordered bases of the $\mathbb{F}_q$ vector space $\mathbb{F}_q^n$), therefore $$\#\operatorname{Aut}(\mathbb{Z}_6 \times \mathbb{Z}_{10} \times \mathbb{Z}_{15}) \\ = \#(\operatorname{GL}(2,2) \times \operatorname{GL}(2,3) \times \operatorname{GL}(2,5)) \\ = \#\operatorname{GL}(2,2) \cdot \#\operatorname{GL}(2,3) \cdot \#\operatorname{GL}(2,5) \\ = 6 \cdot 48 \cdot 480 \\ = 138240.$$