# If A is a finite abelian p-group, then Aut(A) acts transitively on the elements of highest order.

If A is a finite abelian p-group, then Aut(A) acts transitively on the elements of highest order. Could someone show me why this is true?

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@user9077: sorry I meant highest – Mark Mar 8 '12 at 0:55
Have you seen that finite abelian groups are direct sums of cyclic groups? – Jonas Meyer Mar 8 '12 at 1:01
@JonasMeyer: I have read it some where, but have not seen proof. – Mark Mar 8 '12 at 1:02

You really need the fundamental theorem of finitely generated abelian groups to understand this. By that theorem, a finite abelian $p$-group $P$ is isomorphic to a direct sum of cyclic groups $C_1 \oplus \cdots \oplus C_k$, where $C_i$ is cyclic of order $p^{n_i}$ and $n_1 \le n_2 \le \cdots \le n_k$.

Then the maximal order of elements in the group is $p^{n_k}$. There exists $j$ such that $n_i = n_k$ for $j \le i \le k$. In other words, $C_j,C_{j+1},\ldots,C_k$ are the cyclic direct summands of largest order.

Let $g = (g_1,\ldots,g_k)$ be an element of maximal order $p^{n_k}$. Then at least one of the components $g_i$, with $j \le i \le k$ must have order $p^{n_k}$. Let's assume that $g_k$ has that order.

Let $C_k'$ be the cyclic subgroup of $G$ generated by $g$. Then it is routine to check that $C_k'$ has zero intersection with $C_1 \oplus \cdots \oplus C_{k-1}$ and that $C_k'$ and $C_1 \oplus \cdots \oplus C_{k-1}$ generate the group, so we have $G = C_1 \oplus \cdots \oplus G_{k-1} \oplus C_k'$. (I'm being a bit sloppy about the distinction between isomorphism and equality, but that's not improtant here.)

Hence there is an automorphism of $G$ that maps $g$ to the element $(0,0,\ldots,0,1)$ of $G$, and since this is true for any $g$ of maximal order, we get the required transitivity.

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Let $m=p^a$ be the highest order. If the order of $x$ is $m$ and $f\in\mathrm{Aut}(A)$, then $f(x)^m=f(x^m)=f(e)=e$. Moreover, if $e=f(x)^n=f(x^n)$ then $x^n=e$ since $f$ is an automorphism, therefore $m\leq n$. Conclusion: $f(x)$ has order $m$.

The action if transitive since $f^{-1}$ is also in $\mathrm{Aut}(A)$.

It holds for a general $p$-finite group, no necessary abelian.

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I do not understand how this gives a hint of why the action is transitive. You have indicated why the order of $x$ is the same as the order of $f(x)$ for each automorphism $f$. This is true regardless of the order of $x$, and yet the action is not typically transitive on elements of a given order. E.g., in $\mathbb Z_2\oplus \mathbb Z_4$, there is no automorphism sending $(1,0)$ to $(0,2)$. – Jonas Meyer Mar 8 '12 at 3:23
The final assertion is not true. Let $G$ be the direct product of the dihedral group of order 8 and a cyclic group of order 4. The highest order of an element is 4, but the group has both central and non-central elements of order 4, and no group automorphism can take a central element to a non-central element. – Derek Holt Mar 8 '12 at 4:30