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?
 A: 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.
A: 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.
