order of automorphism group of an abelian group of order 75 Let $G$ be an abelian group of order $75=3\cdot 5^{2}$. Let $Aut(G)$ denote its group of automorphisms. Find all possible order of $Aut(G)$.
My approach is to first study its Sylow 5-subgroup. Since $n_{5}|3$ and $n_{5}\equiv 1\pmod{5}$, $n_{5}=1$. So $G$ has a unique Sylow 5-subgroup, denote $F$. By Sylow's Theorem, $F$ is characteristic. Then $\forall \sigma\in Aut{G}$,  $\sigma(F)=F$. Define the canonical homomorphism $Aut(G)\rightarrow Aut(H)$. So $|Aut(G)|=\# (\text{Aut(G) that fixes H pointwise})\times (\text{image of homomorphism})$. 
Since the image of the defined homomorphism is a subgroup of $Aut(F)$, then its order is a factor of $Aut(F)=20$. I'm not sure how to compute the number of $Aut(G)$ that fixes $H$. My understanding is this: Since we leave $H$ fixed, all that left to be permuted are the 25 Sylow 3-subgroup. Since each Sylow 3-subgroup is cyclic, it has 2 automorphisms. So altogether $|\text{Aut(G) that fixes H pointwise}|=25\times 3=75$. And all possible order of $Aut(G)$ is $75\cdot x$ where $x$ divides 20. This seems incorrect. Could someone please help me?
 A: Sylow theory is generally not useful for Abelian groups since we already know so much about their structure, and since every Abelian group has a unique $p$-Sylow subgroup when it exists.
Using the classification theorem, we have that $|{G}| = 75$ implies $G \cong \mathbb{Z}/3\mathbb{Z} \times (\mathbb{Z}/5\mathbb{Z})^2$ or $G \cong \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/25\mathbb{Z} \cong \mathbb{Z}/75 \mathbb{Z}$.
Now we use standard combinatorical methods to count the automorphisms in each case.
In the first case we must send $(1,0,0)$ to an element of order $3$, giving $2$ choices. We must send $(0,1,0)$ to an element of order $5$, giving $24$ choices. We must send $(0,0,1)$ to an element of order $5$ so that our function is surjective, giving $24 - 4 = 20$ choices. That is we are subtracting off the elements in the subgroup generated by the image of $(0,1,0)$. After defining our function on a basis, we can extend uniquely to a homomorphism on the group. This gives $20 \cdot 24 \cdot 2 = 960$ automorphisms. This work is made easier by noting $\text{aut}(H) \times \text{aut}(K) \cong \text{aut}(H \times K)$ when $H$ and $K$ are abelian groups of relatively prime orders. 
In the second case we must simply pick a generator, giving $\phi(75)$ automorphisms, where $\phi$ is the Euler totient function.
A: $G$ is abelian, so Sylow subgroups are characteristic. Hence $G$ is product of characteristic Sylow  subgroups, say $H_3$ and $H_5$. Then $Aut(G)\cong Aut(H_3)\times Aut(H_5)$. 
$H_3$ is cyclic, whose automorphism group is well known.  $H_5$ is either cyclic or $Z_5\times Z_5$. The automorphism groups in both cases in well known: either $Z_5\times Z_4$ or $GL_2(Z_5)$, the group of $2\times 2$ invertible matrices over $Z_5$ (think on it, or see some book, or on-line reference). You can obtain the order of $Aut(G)$ easily.
