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The automorphism group of a cyclic group $\mathbb{Z}_n$ has an order given by Euler's totient function $\phi(n)$, where $\phi(n)$ is the number of generators of $\mathbb{Z}_n$. The answer in this question says that this is because generators must map to generators, presumably as an isomorphism (and hence an automorphism) preserves element order.

Consider some groups such all non-identity elements have equal order:

  • The Klein-4 group $\mathbb{Z}_2 \times \mathbb{Z}_2$ has the identity element and 3 elements of order 2. Taking permutations of these 3 elements gives an automorphism group of order $3! = 6$, which is the correct order.

  • The Elementary Abelian group $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$ has the identity element and 7 elements of order 2. Taking permutations of these 7 elements gives an automorphism group of order $7! = 5040$, but the correct order is $168$.

  • The group $\mathbb{Z}_3 \times \mathbb{Z}_3$ has the identity element and 8 elements of order 3. Once again taking permutations give the automorphism group an order of $8! = 40320$, but the correct order is $48$.

Is there a way to compute the order of the automorphism groups from element order alone?

Some examples use the result ${\rm Aut}(C_p\times C_p)\simeq{\rm GL}_2(\mathbb Z_p)$, but interested to see if it can be computed using the element order alone.

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    $\begingroup$ You need to map generators to generators and not just take a permutation of the elements. $\endgroup$ – Dietrich Burde Jun 8 at 16:41
  • $\begingroup$ But why can't say any element of order 2, map to another element of order 2? $\endgroup$ – pymekrolimus Jun 8 at 17:59
  • $\begingroup$ Because (i) they may get forced by what happens to other elements of order $2$; the Klein 4-group is special because you don't have enough "wiggle" room so the two values coincide; and (ii) because not every element of order $2$ need be in a generating set. (This is more obvious with groups of exponent $p$ for odd $p$ that are not abelian). $\endgroup$ – Arturo Magidin Jun 8 at 18:01
  • $\begingroup$ If you just map each element independently, you won't always obey the homomorphism equation $\phi(xy) = \phi(x)\phi(y)$. $\endgroup$ – aschepler Jun 8 at 23:24
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A homocyclic abelian group with all elements of prime order are the easiest case: that would be groups of the form $C_p^n$ with $p$ prime and $n\gt 0$. These are $n$-dimensional vector spaces over the field of $p$ elements, and so their automorphisms are given precisely by $\mathrm{GL}_n(\mathbb{F}_p)$. The order is known to be $$(p^n-1)(p^n-p)\cdots (p^n-p^{n-1})$$ which can be verified as follows: an automorphism is determined by what it does to the standard ordered basis. There are $p^n-1$ choices of where to send the first vector (any nonzero vector); then there are $p^n-p$ choices for the second (any vector not a multiple of the image of the first); then $p^n-p^2$ choices for the second one (any vector not in the span of the images of the first two), etc.

But even for prime order the result is very different once you leave homocyclic groups. The nonabelian group of order $p^3$ and exponent $p$ (also known as the Heisenberg group over $\mathbb{F}_p$), for example, has the same number of elements of order $p$ as the homocyclic group $C_p^3$; but while the latter has automorphism group of order $(p^3-1)(p^3-p)(p^3-p^2)$, the former is much more delicate: not every element of order $p$ can be the image of a generator.

The group can be described by the presentation $$\langle x,y,z\mid x^p=y^p=z^p=1, yx=xyz, xz=zx, yz=zy\rangle.$$ Every automorphism is completely determined by what it does to $x$ and to $y$. Now, $x$ will map to $x^ay^bz^c$, $y$ maps to $x^ry^sz^t$, where $a,b,c,r,s,t$ are integers modulo $p$, we must have $as-rb\not\equiv 0\pmod{p}$, but there are no conditions on $c$ and $t$. That means that here you get a group that has order $(p^2-1)(p^2-p)p^2$, since the $4$-tuple $(a,b,r,s)$ must correspond to an invertible $2\times 2$ matrix with coefficients in $\mathbb{Z}/p\mathbb{Z}$, and the values of $c$ and $t$ are free.

For example, for $p=3$, the homocyclic group $C_3^3$ will have $26$ elements of order $3$, and an automorphism group of order $(3^3-1)(3^3-3)(3^3-9) = 11,232$, whereas the Heisenberg group also has $26$ elements of order $3$, but an automorphism group of order $432$, because all the "action" is happening with two elements, rather than three linearly independent elements as in the homocyclic case.

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The situation is more delicate than you were hoping it to be.

In the very simplest case, which all your examples fall into, namely groups of form $(\Bbb Z_p)^n$, you should think of them as vector spaces $V$ over the field $\Bbb F_p=\Bbb Z/p\Bbb Z$, dimension $n$. Now, any automorphism of such a group is also an $\Bbb F_p$-linear automorphism, thus determined by an $n$-by-$n$ matrix, which must be nonsingular.

Let $\{v_1,\cdots,v_n\}$ be a basis of your $n$-dimensional $\Bbb F_p$-space. You send $v_1$ to any nonzero element of $V$, $p^n-1$ choices. Now send $v_2$ to any element of $V$ not in the span $\langle v_1\rangle$, thus $p^n-p$ new choices. Now send $v_3$ to any element of $V$ not in the span $\langle v_1,v_2\rangle$, $p^n-p^2$ new choices. Continue, and you see that you have $(p^n-1)(p^n-p^2)\cdots(p^n-p^{n-1})$ possibilities for your nonsingular matrix, and that’s the answer.

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