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It is not too hard to find examples of finite groups which have fewer automorphisms than one of their subgroups. For example $\mathcal D_4 \times \mathbb Z/2\mathbb Z$ (where $\mathcal D_4$ is the dihedral group of order $8$) has at most (in fact exactly) $64$ automorphisms but contains a subgroup isomorphic to $\left(\mathbb Z/2\mathbb Z\right)^3$, which has $\operatorname{GL}_3(\mathbb F_2)$ as automorphism group, with $168$ elements.

Every example I found was non-abelian, so I wondered if a finite abelian group can have less automorphism than one of its subgroup? Hillar and Rhea - Automorphisms of finite Abelian groups gives the number of automorphisms of an abelian $p$-group, but it's not clear to me if that function is non-decreasing with cardinality.

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  • $\begingroup$ A good candidate is something like $(\mathbb{Z}/2\mathbb{Z})^3$ inside $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/8\mathbb{Z}$ $\endgroup$
    – Noah Snyder
    Commented Mar 2, 2023 at 20:56
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    $\begingroup$ @NoahSnyder That doesn't work, they have automorphism groups of order $168$ and $2048$. The closest I can find is $({\mathbb Z}/2)^n$ inside ${\mathbb Z}/4 \times({\mathbb Z}/2)^n$, but the second one seems to have very slightly larger automorphism group. $\endgroup$
    – Derek Holt
    Commented Mar 2, 2023 at 21:03
  • $\begingroup$ So probably there aren't any? But the fact that the reduction mod 2 map (see Thm 3.6) has both an interesting kernel and an interesting cokernel in that example (whose sizes you then need to compare) suggests the proof will be kind of messy. $\endgroup$
    – Noah Snyder
    Commented Mar 2, 2023 at 21:16
  • $\begingroup$ @DerekHolt If the subgroup is a direct summand, then the whole group will definitely have at least as many automorphisms, since you can just act on one of the factors fixing the other. $\endgroup$
    – Wojowu
    Commented Mar 2, 2023 at 21:18
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    $\begingroup$ Sorry, typo, I meant "inside ${\mathbb Z}/4 \times ({\mathbb Z}/2)^{n-1}$". $\endgroup$
    – Derek Holt
    Commented Mar 2, 2023 at 21:56

1 Answer 1

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From the Hiller-Rhea formula $$|\operatorname{Aut} H_p| = \prod_k (p^{d_k} - p^{k-1}) \prod_j (p^{e_j})^{n-d_j} \prod_i (p^{e_i-1})^{n-c_i+1},$$ given an abelian $p$-group of type $p^{e_1}\cdots p^{e_n}$ where $1 \leq e_1 \leq \cdots \leq e_n$, you can try to increment one of the exponents $e_l$ by 1, and see how things change. Here, $d_k = \max \{l | e_l = e_k\}$ and $c_k = \min \{ l | e_l = e_k \}$.

Case 1: $e_{l-1} < e_l$ and $e_l + 1 < e_{l+1}$. The only changes to the formula are $(p^{e_l})^{n-d_l}$ and $(p^{e_l-1})^{n-c_l+1}$, and yield a strict increase.

Case 2: $e_l$ is isolated, but $e_l + 1$ is equal to $m$ other values. Then, $d_l$ jumps by $m$, and $c_i$ decrease by $1$ for $m$ entries. We end up adding an exponent of $2n - 2l + 1 + \epsilon > 0$, where $\epsilon$ is a way to ignore the $p^{k-1}$ in the first product.

Case 3: $e_l$ is equal to $m$ other values, but $e_l+1$ is isolated. Then, $d_i$ drops by $1$ for $m$ entries, and $c_l$ jumps by $m$. We add an exponent of $2n-2l+1 - \epsilon > 0$.

Case 4: $e_l$ is equal to $m$ other values, and $e_l+1$ is equal to $r$ other values. Then, $d_i$ drops by $1$ for $m$ entries, $d_l$ jumps by $r$, $c_i$ decrease by $1$ for $r$ entries, and $c_l$ jumps by $m$. We add an exponent of $2n-2l+1 - \epsilon > 0$.

In conclusion, a subgroup of a finite abelian group always has strictly smaller automorphism group (Edit: with the exception of a group of order $2$ times an odd number, and the subgroup of index 2, where we have equality). The minimal case is when $e_1 = \cdots = e_n$. Then, incrementing $e_n$ multiplies the order by $\frac{p^n(p-1)}{p^n-1}$, which approaches 1 when $p=2$ and $n$ is large.

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    $\begingroup$ The point is that you produce all subgroups this way. The precise statement is that if we set $f_i = e_{n-i}$ and $f_i = 0$ for $i > n$, then a $p$-group $H'$ of type $(f'_1,f'_2,\ldots)$ appears as a subgroup of $H$ if and only if $f'_i \leq f_i$ for all $i$. This is surely well-known, but requires a little argument. (The index $p$ case then corresponds to lowering one index by $1$.) I suppose one case that's missing from your answer is where $f'_n$ becomes $0$, i.e. you drop a factor, but this case is trivial. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Mar 3, 2023 at 11:01
  • $\begingroup$ My problem was exactly what R. van Dobben de Bruyn noticed, it's probably true but I'm missing the proof that for a subgroup $f_i' \leq f_i$ for all $i$. $\endgroup$
    – A. Bailleul
    Commented Mar 3, 2023 at 14:42
  • $\begingroup$ @A.Bailleul One argument is as follows: given an abelian group $H$, write $d(H)$ for the minimal number of generators of $H$. Since $\mathbf Z$ is a PID, any subgroup $H'\subseteq H$ satisfies $d(H')\leq d(H)$. But if $H\cong \prod_{i=1}^n\mathbf Z/p^{f_i}\mathbf Z$, then $d(p^iH)=\#\{j\ |\ f_j>i\}$. Thus the inequality $d(p^iH')\leq d(p^iH)$ gives $\#\{j\ |\ f'_j>i\}\leq\#\{j\ |\ f_j>i\}$ for all $i$, which means exactly that $f'_i\leq f_i$ for all $i$ since the $f_i$ and $f'_i$ are sorted in decreasing order. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Mar 3, 2023 at 16:36

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