# 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.

• Does this work? Since the $G_i$ are characteristic, we get a homomorphism $G \rightarrow \oplus_{i = 1}^t \operatorname{Aut}(G_i / G_{i-1})$ which has as its kernel the subgroup $N$. Here $\operatorname{Aut}(G_i / G_{i-1})$ is cyclic of order $p-1$, so you get $|G| = (p-1)^t p^r$ for some $r$. – spin Dec 5 '17 at 19:09
• @spin: Yes, that's very nice! Note however that your conclusion is incorrect, and is weaker than the full result I'd like: the exponent of $(p-1)$ should be $k$ (the number of summands of $G$). – Steve D Dec 5 '17 at 20:05
• Oh right, you are correct. Then I don't know. Maybe it is worth noting that the result is not true if we do not have strict inequalities $n_i < n_{i+1}$, so that needs to be used somehow. – spin Dec 5 '17 at 20:25
• @spin: The existence of the characteristic series depends on the strict inequalities (it isn't true for $\mathbb{Z}_2\oplus\mathbb{Z}_2$, for example). – Steve D Dec 5 '17 at 20:26

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.