Are all power automorphisms of a finite abelian group of the form $f(x)=x^n$?

Question

An example of a universal power automorphism, that is automorphisms of the form $f(x)=x^n$, is easy to find in an abelian group. Any power $n$ will work as long it is relatively prime to the orders of all elements of the group.

More generally, a power automorphisms sends any $g \in G$ to a power of $g$. Equivalently, it maps all subgroups $H$ of $G$ to H. My hunch is that any power automorphism of an abelian group must be a universal power automorphism. Is this true?

(In general, power automorphisms that aren't universal power automorphisms do exist for finite groups. See my question here. The example in that case is non-abelian, however.)

Answer 1

Every finitely generated abelian group has a decomposition

$$ \mathbb{Z}/n_1\oplus\mathbb{Z}/n_2\oplus\cdots\oplus\mathbb{Z}/n_k $$

for some moduli $n_1,n_2,\cdots,n_k$.

Let $\phi$ be a so-called power automorphism. Use $g_i$ for the coordinate basis, so $g_1=(1,0,\cdots,0)$ for example. Then $\phi(g_i)=e_ig_i$ for $i=1,\cdots,k$ for some numbers $e_1,\cdots,e_k$. (In multiplicative notation, these would be exponents.) So $\phi$ acts "diagonally" via the formula

$$ \phi(x_1,x_2,\cdots,x_n)=(e_1x_1,e_2x_2,\cdots,e_nx_n). $$

Since $\phi$ is a power automorphism, we may write

$$ \begin{array}{ll} \phi(1,1,\cdots,1) & =e(1,1,\cdots,1) \\ & =(e,e,\cdots,e) \end{array} $$

for some integer $e$ and also we may write

$$ \phi(1,1,\cdots,1)=(e_1,e_2,\cdots,e_k). $$

Therefore, $e\equiv e_i \mod n_i$ for each $i=1,\cdots,k$ and so we have

$$ \phi(x_1,\cdots,x_n)=(e_1x_1,\cdots,e_kx_k)=(ex_1,\cdots,ex_k) $$

i.e. $\phi(x)=ex$ (or in multiplicative notation, $\phi(x)=x^{\large e}$), Q.E.D.

Note it's okay for any of $n_1,\cdots,n_k$ to be $0$, so $G$ doesn't need to be finite.

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It's not true for infinite torsion abelian groups. In particular, we can get "power-like" automorphisms that are not actually integer power maps. For instance, in the $p$-primary component of the additive factor group $\mathbb{Q}/\mathbb{Z}$, the so-called Prufer $p$-group which is the additive group of $\mathbb{Z}[1/p]$ mod the subgroup $\mathbb{Z}$, we can multiply by a $p$-adic integer which doesn't need to be a rational integer, or even rational at all.