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?

  • 4
    $\begingroup$ Start by using the Fundamental Theorem of Abelian Groups to identify all the possible abelian groups of order $75$. You do not need Sylow subgroups per se as the FTAG gives these possible groups explicitly. $\endgroup$ – hardmath Jan 3 '16 at 3:40

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.

  • $\begingroup$ "Sylow theory is generally not useful for Abelian groups" (can you elaborate this for me?) In fact Sylow subgroups in abelian group are characteristic; they make things much simple, isn't it? $\endgroup$ – Beginner Jan 3 '16 at 10:30
  • $\begingroup$ @EricThoma Could you explain a bit more why sending (0,1,0) to an element of order 5 gives 24 choices? $\endgroup$ – Orca_1 Jan 3 '16 at 18:19
  • $\begingroup$ @MarshalKurosh I just mean that the classification theorem is generally stronger, and anything that can be deducted from Sylow theory is much easier deducted from the classification theorem. At least intuitively, Sylow theory should be trivial in abelian groups since conjugation is trivial in abelian groups. That all Sylow $p$-subgroups are conjugate is a useful fact, but I contend that this is still easier with the classification theorem. $\endgroup$ – Eric Thoma Jan 3 '16 at 22:10
  • $\begingroup$ @JiayuanW The factors of $24$ and $20$ are from considering automorphisms of $(\mathbb{Z}/5\mathbb{Z})^2$. We must send $(0,1,0)$ to an element of order $5$ in this group, i.e. any non-identity element. There are $25 - 1 = 24$ such elements. $\endgroup$ – Eric Thoma Jan 3 '16 at 22:14

$G$ 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.

  • $\begingroup$ Do you mean all Sylow subgroups in an Abelian group are characteristic? $\endgroup$ – Orca_1 Jan 3 '16 at 18:55
  • $\begingroup$ Also, how should one compute the order of $GL_{2}(Z_{5})$? I understand that there are 5 choices for every entry in the $2\times 2$ matrix. But some choices should be subtracted from $5^{4}$ in order to make the matrix invertible. Exactly how would one count them? $\endgroup$ – Orca_1 Jan 3 '16 at 19:01
  • 1
    $\begingroup$ Yes-in finite abelian groups, Sylow subgroups are characteristics. For computation of order of $GL_2(5)$, note that a $2\times 2$ matrix over $Z_5$ is invertible if and only if its columns are independent. First column $[a,b]$ has $5.5$ choices, but we exclude $[0,0]$. Second column $[c,d]$ has (initially) $5.5$ choices but it should be independent of $[a,b]$, so it shouldn't be scalar multiple of $[a,b]$; so exclude the $5$ scalar multiples. Thus, $[a,b]$ has $(5^2-1)$ and $[c,d]$ has $(5^2-5)$ choices; so $|GL_2(5)|=(5^2-1)(5^2-5)$. $\endgroup$ – p Groups Jan 4 '16 at 2:46

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