Order of the group of automorphisms of $\mathbb Z_{15} \times\mathbb Z_3$ 
What is the order of group of automorphisms of $\mathbb Z_{15} \times\mathbb Z_3$? 

If $15$ and $3$ were relatively prime I would know how to compute the order, but since they are not, I don't know how to do it. 
 A: Well, note that since $\mathbb{Z}_5$ and $\mathbb{Z}_3^2$ are coprime in orders, that 
$$|\text{Aut}(\mathbb{Z}_{5}\times\mathbb{Z}_{3}^2)|=|\text{Aut}(\mathbb{Z}_5)|\times|\text{Aut}(\mathbb{Z}_3^2)|$$ 
Now, $\text{Aut}(\mathbb{Z}_5)\cong \mathbb{Z}_4$ and $\text{Aut}(\mathbb{Z}_3^2)\cong \text{GL}_2(\mathbb{Z}^3)$. It's common knowledge that 
$$|\text{GL}_n(\mathbb{F}_q)|=(q^n-1)(q^n-q)\cdots(q^n-q^{n-1})$$
So, 
$$|\text{GL}_2(\mathbb{Z}_3)|=(3^2-1)(3^2-3)=48$$
So, the order is $48\cdot 4=192$.
A: We know that $\mathbb{Z}_{15}\times\mathbb{Z}_3\cong\mathbb{Z}_5\times\mathbb{Z}_3\times\mathbb{Z}_3 = G$. All of those are cyclic groups, thus a group homomorphism $G\rightarrow H$ is completely determined by the image of the three generators.
The generator of $\mathbb{Z}_5$ can only be mapped to one of the non-identity elements of $\mathbb{Z}_5$ (else we will not get an automorphism), so we have $4$ possibilities.
The generator of the first copy of $\mathbb{Z}_3$ can only be mapped to an element of $\mathbb{Z}_3\times\mathbb{Z}_3$ of order $3$, there are $8$ of them (all elts of $\mathbb{Z}_3\times\mathbb{Z}_3$ have order $3$, except for the identity).
The generator of the second copy of $\mathbb{Z}_3$ will be mapped to any remaining non-identity element. There are $6$ of them.
Thus we get $4\cdot8\cdot6=192$ possible automorphisms.
