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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.

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    $\begingroup$ 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 $r$. $\endgroup$
    – spin
    Dec 5, 2017 at 19:09
  • $\begingroup$ @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 $G$). $\endgroup$
    – Steve D
    Dec 5, 2017 at 20:05
  • $\begingroup$ 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. $\endgroup$
    – spin
    Dec 5, 2017 at 20:25
  • $\begingroup$ @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). $\endgroup$
    – Steve D
    Dec 5, 2017 at 20:26

2 Answers 2

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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.

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The accepted answer constructs a series of subgroups, but doesn't explain in detail why the series is characteristic. It hints there is a more insightful argument than the calculations given here -- maybe somebody can post that. Also, we point out where in the accepted answer it is necessary that the series is characteristic.

In a characteristic series each term of the series is a characteristic subgroup of the whole group. (Robinson defines the term later in Chapter 3, and there gives an ambiguous definition that can be read as a strongly characteristic series. Thanks to @Steve D for pointing out the difference.)

The series given contains duplicate terms. Since each non-trivial factor has size $p$, the series has $t=\sum_{i=1}^k n_i$ non-trivial factors, while the construction has $kn_1>t$ terms.

The series is characteristic

To aid calculations we introduce new notation. Rewrite the order constraints in the accepted answer as $G_m^i=\oplus_{j=1}^k G_j^{i,m}$ with $$ G_j^{i,m} = p^{r_j^{i,m}}\mathbb{Z}_{p^{n_j}} $$ where $$ r_j^{i,m}=\max(0,n_j-m+\delta_{j>i}) $$ and $$ \delta_{j>i}=\left\{\begin{array}{cl} 1 & j>i \\ 0 & j\leq i \end{array}\right. $$

For $i>0$ we have $G_j^{i-1,m}=G_j^{i,m}$ for $j\neq i$ and $$ G_i^{i-1,m} = \left\{\begin{array}{cl} pG_i^{i,m} & m\leq n_i \\ G_i^{i,m} & m>n_i \end{array}\right. $$ Similarly we have $$ G_j^{k,m-1} = \left\{\begin{array}{cl} pG_j^{1,m} & j=1 \\ G_j^{1,m} & j>1 \end{array}\right. $$

We can take the generator $h^i_m$ to be $$ h^i_m=(0,\ldots,p^{n_i-m},\ldots,0) $$

Lemma If $s_1, s_2\geq 0$ and $\eta:\mathbb{Z}_{p^{s_1}}\rightarrow \mathbb{Z}_{p^{s_2}}$ is a homomorphism then for some $c\geq 1, u\in\mathbb{Z}$ and all $k\in\mathbb{Z}$ we have $\eta(k)=ckp^{u}\bmod p^{s_2}$ with $p\nmid c$ and $\max(0,s_2-s_1)\leq u\leq s_2$.

Proof:
The homomorphism $\eta$ is determined by its value on a generator $\eta(1)=cp^{u}$, where $p\nmid c$ and $0\leq u\leq s_2$. For $\eta$ to be well-defined requires $p^{s_2}\,|\,\eta(p^{s_1})=p^{s_1}\eta(1)$ or $s_2\leq u+s_1$.

Robinson's preceding exercise 1.5.10 suggests writing an element $\alpha\in\mbox{Aut }(G)$ as a $k\times k$ invertible matrix of homomorphisms. If $\alpha\in\mbox{Aut }(G)$ and $y=\alpha(x)$, then $y_j=\sum_{l=1}^k\alpha_{jl}x_l$ where the matrix element $\alpha_{jl}:G\rightarrow G$ is a multiplication $\alpha_{jl}=c_{jl}p^{u_{jl}}$ with $u_{jl}\geq \max (0, n_j-n_l)$.

Proposition The series is characteristic.

Proof:
We proceed by induction on the terms of the series. The base of the induction is $\alpha(\{0\})=\{0\}$ for $\alpha\in\mbox{Aut}(G)$.

For later terms, we need to show that if $m\leq n_i$ then \begin{align*} \alpha(h^i_m) &\in G_m^i\\ c_{ji}p^{u_{ji}+n_i-m} &\in p^{r^{i,m}_j}\mathbb{Z}_{p^{n_i}}\mbox{ for }1\leq j\leq k\\ u_{ji}+n_i-m &\geq \max(0,n_j-m+\delta_{j>i}) \end{align*} This follows from the lemma by separately considering the cases $j=i$, $j<i$ and $j>i$.

Where the characteristic series is used

That the series is characteristic is needed in two places in the accepted answer.

The subgroup $N$ is defined as the kernel of a homomorphism $\mbox{Aut}(G)\rightarrow\prod_i\mbox{Aut}(G_i/G_{i-1})$. This homomorphism is defined only when $G_i$ and $G_{i-1}$ are characteristic subgroups of $G$.

Later, it is argued that the "diagonal endomorphisms" $\phi_i$ of an automorphism $\phi\in\mbox{Aut}(G)$ are themselves automorphisms. Choose $j$ so that $G_j=G^i_{n_i}$ (e.g., $j=k(n_i-1)+i$). Let $\pi_i:G\rightarrow\mathbb{Z}_{p^{n_i}}$ be the $i^\text{th}$ projection. Then, since $G_j$ and $G_{j-1}$ are characteristic subgroups of $G$, \begin{align*} h^i_{n_i} &\in G_j\setminus G_{j-1}\\ \phi(h^i_{n_i}) &\in G_j\setminus G_{j-1}\\ \pi_i(\phi(h^i_{n_i})) &\in \pi_i(G_j)\setminus \pi_i(G_{j-1})\\ \phi_i(\pi_i(h^i_{n_i})) &\in \mathbb{Z}_{p^{n_i}}\setminus p\mathbb{Z}_{p^{n_i}} \end{align*} so $\phi_i$ is an automorphism.

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  • $\begingroup$ I think in general there is no strongly characteristic series for this group with factor size $p$. $\endgroup$ Feb 19 at 14:06

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