Order of automorphism group of abelian group In Derek Robinson's A Course in the Theory of Groups, exercise 1.5.13 states:

Let $G=\mathbb{Z}_{p^{n_1}}\oplus\cdots\oplus\mathbb{Z}_{p^{n_k}}$, where $n_1<n_2<\cdots<n_k$. Prove there exists a chain of characteristic subgroups $1=G_0<G_1<\cdots<G_t=G$ such that $[G_{i+1}:G_i]=p$ and $t=\sum n_i$. Deduce that $|Aut(G)|=(p-1)p^r$ for some $r$.

Now, this exercise is wrong.  What is true, however, is that
$$|Aut(G)|=(p-1)^kp^r$$
for some $r$.  Is there an elementary way to prove this? It's pretty easy to see that if $\alpha\in Aut(G)$ fixes pointwise the quotients $G_{i+1}/G_i$, then it has order a power of $p$. If $N\lhd Aut(G)$ is the subgroup of all such $\alpha$, then for every $xN\in Aut(G)/N$, $(xN)^{p-1}=1$. That is, $Aut(G)/N$ has exponent $p-1$. But I don't see a way to show $|Aut(G)/N|=(p-1)^k$.
Of course, it is entirely possible such an easy proof does not exist. But that makes me wonder what the point of Robinson's exercise is.
 A: Here's an approach that I think works. A little bit much for an exercise, but the key ideas are fairly simple.
Preliminaries
I'll change things up and write $G=\mathbb{Z}_{p^{n_1}}\oplus\cdots\oplus\mathbb{Z}_{p^{n_k}}$, with $n_1>n_2>\cdots>n_k$.  This is backwards from how I originally wrote it, but it's easier to read left-to-right, and makes the indexing a little simpler below. I'll also use the additive notation for the group operation in $G$.
I'll write generic elements of $G$ as $(g_i)$, where the $i$th component belongs to the $i$th factor, $\mathbb{Z}_{p^{n_i}}$. We can define the characteristic series of $G$ via subgroups $G^i_m$.  A generic element of $G^i_m$ looks like $(a_j)$, where
\begin{align}
|a_j| &\le p^m&&j\le i\\
|a_j| &< p^m&&j> i
\end{align}
Basically, we're building "horizontally".  We go left-to-right, collecting all elements of order at most $p$.  Then, a carriage return, and then we go left-to-right, collecting all elements of order at most $p^2$, and so on.
[The reason the $G^i_1$ are characteristic is because they can be written in terms of "all elements of order $p$, with sufficiently many roots".  The $G^i_2$ are then $p$th roots of the $G^i_1$, etc.]
To be explicit, the series goes
$$ 1 < G^1_1 < G^2_1 < \cdots < G^k_1 < G^1_2 < G^2_2 < \cdots < G$$
Finally, I'll let $h_{i,m}$ denote a once-and-for-all chosen generator of a factor of the characteristic series.  That is, $h_{1,m}$ generates $G^1_m/G^k_{m-1}$, and $h_{i,m}$ generates $G^i_m/G^{i-1}_m$ for $i>1$. Note that we can pick 
$$ h_{i,m} = (0, \ldots,h_i, 0, \ldots)$$
That is, only the $i$th component of $h_{i,m}$ is nonzero.   Here, $|h_i|=p^m$. Note that the coset for $h_{i,m}$ in the factor group is
$$ h_{i,m}G^{i-1}_m = (a_1, \ldots, h_it, b_1, \ldots)$$
where $|a_j|\le p^m$, $|b_j|<p^m$, and $|t|<p^m$.
Normal Subgroup
Let $A=Aut(G)$.  We define the subgroup $N$ to be all elements of $A$ that act trivially on each factor group of the characteristic series. 
Lemma: If $\alpha\in N$, then $|\alpha|$ is a power of $p$.
Proof: By induction.  Pick $n=p^s$ high enough, such that $\alpha^n$ acts trivially on most of the series (all but $G$), and then $\alpha^n(h_{1,n_k})=h_{1,n_k}x$, where $\alpha^n(x)=x$. If $q=|x|$, then $\alpha^{nq}$ acts as the identity on $G$. $\blacksquare$
So $N$ is a $p$-subgroup of $A$.  It's easy to show directly that it is normal, or we can note it is the kernel of the map $A\rightarrow\prod Aut(\text{factor})$, the target group being the direct product of the automorphism groups of each series factor. Since each of these automorphism groups is cyclic of order $p-1$, we see that $p$ does not divide $|A/N|$, and thus $N$ is the normal Sylow $p$-subgroup of $A$.
Complement Subgroup
For each $\mathbb{Z}_{p^{n_i}}$, there is an automorphism $\psi_i$ of order $p-1$. We can treat $\psi_i$ as an element of $A$, that acts trivially on all other factors.  Since each $\psi_i$ has order $p-1$, and the various $\psi_i$ commute with each other, $K=\langle\psi_i\rangle$ is an abelian subgroup of $A$, with $|K|=(p-1)^k$.
Semidirect Product
By order considerations, $N\cap K=1$.  So if we show $A=NK$, then we have written $A$ as a semidirect product of $N$ and $K$, and in particular, 
$$ |A| = |N||K| = p^r(p-1)^k$$
So for now on, let $\phi\in A$ be an arbitrary automorphism.
First, note that, if we consider $Aut(\mathbb{Z}_{p^{n_i}})$ as a subgroup of $A$, then $Aut(\mathbb{Z}_{p^{n_i}})\subset NK$. This is because we can write any such automorphism as $\gamma\psi_i^c$, with $\gamma$ having $p$-power order [essentially, this all boils down to $Aut(\mathbb{Z}_{p^{n_i}})$ having order $p^{n_i-1}(p-1)$].
Also note that $h_{i, n_i}$ is a generator of $\mathbb{Z}_{p^{n_i}}$, and all the non-trivial cosets in the corresponding series factor group are generated by elements of $G$ whose $i$th coordinate is a generator of $\mathbb{Z}_{p^{n_i}}$. So $\phi(h_{i,n_i})$ must also have a generator of $\mathbb{Z}_{p^{n_i}}$ in the $i$th component.  All this is to say, the composition 
$$ \mathbb{Z}_{p^{n_i}}\hookrightarrow G\xrightarrow{\phi} G\twoheadrightarrow\mathbb{Z}_{p^{n_i}} $$
is an automorphism, which I'll denote $\phi_i$. Let $\beta=(\phi_1, \ldots, \phi_k)$. Because $\beta\in NK$, all we have to show is $\beta^{-1}\phi\in NK$.
Finish
Something even stronger is true.  Note that 
$$ \phi(h_{i,m}) = (a_1, \ldots,\overline{h_i}, b_1, \ldots) $$
Now by the "single component" definition of $h_{i,m}$, and by the component-wise definition of $\beta$, we have
$$ \beta^{-1}\phi(h_{i,m}) = (\ldots, h_i, \ldots) $$
which is in the same factor coset of $h_{i,m}$.  Since $h_{i,m}$ generates its corresponding factor group, $\beta^{-1}\phi$ acts pointwise trivially on this factor group.  Since $i$ and $m$ were arbitrary, we see that $\beta^{-1}\phi\in N$, and we're done.
