Herstein's Topics in Algebra provides a proof of the fundamental theorem of finite abelian groups, that is, every finite abelian group is the direct product of cyclic groups.

In an earlier exercise, it is proved that any finite abelian group is isomorphic to the direct product of its Sylow subgroups. In light of this, the proof of the theorem deals only with the case for abelian groups of order $p^n$ (for $p$ prime).

Below is the exposition/sketch of the proof that Herstein offers before diving in to the details:


I highlighted in red and blue the parts I don't understand.

Red: Why is this the case? So we assume each $x\in G$ can uniquely written as $x=a_1^{\alpha_1}\cdots a_k^{\alpha_k}$, with $|a_1|=p^{n_1}$, but at first glance $x$ consists of other factors too, so why is it that $|x|\leq p^{n_1}$ necessarily?

Blue: Here we assume $a_1\in G$ has maximal order, and $\langle a_1\rangle=A_1<G$. What is meant by "$a_2$ maps into an element of highest order in $G/A_1$"? Is this talking about the canonical homomorphism $a_2\mapsto A_1a_2$ (where $a_2$ maps to the coset $A_1a_2$)? If so, how do we know this coset has the highest order in $G/A_1$?

(Perhaps the statement in blue can be answered if I have a better understanding of what is meant by the statement in red.)

Thank you in advance!


For the red statement: what does $x^{p^{n_1}}$ equal?

For the blue statement: Yes, they are talking about the canonical morphism $G \to G/A_1$. You can show that $a_2$ has order $p^{n_2}$ by showing that all its powers lie in distinct cosets. Then show that the order is maximal by showing that $x^{p^{n_2}}$ is in $A_1$ for any x.

  • $\begingroup$ To the OP: as a further hint for the red statement, keep in mind that $n_1\ge n_2\ge ...$ - what can you say about $p^c$ versus $p^d$ for $c\ge d$? What does this have to do with the order of $x$? $\endgroup$ – Noah Schweber May 17 '20 at 20:07
  • $\begingroup$ @Jacob FG Thanks for the response - the red statement is clear to me now. I must have ignored that $n_1\geq\dots\geq n_k$. Could you provide a little more details on the blue statement, however? I hope I'm understanding the goal correctly: So we start off knowing only that $a_1\in G$ has maximal order. Then are we trying to find some $a_2$ so that the coset $A_1a_2$ has maximal order in $G/A_1$? And after finding such an $a_2$, how do we show that $|a_2|=p^{n_2}$, if we don't know a priori what $n_2$ is? Or maybe I'm understanding the premises incorrectly... $\endgroup$ – buffle May 17 '20 at 20:08
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    $\begingroup$ @NoahSchweber Yes, that's exactly what I had in mind, I just didn't account for the fact that $p^{n_1}\geq p^{n_2}$. Thanks! $\endgroup$ – buffle May 17 '20 at 20:10
  • $\begingroup$ @buffle The premise is that $a_i$ has order $p^{n_i}$ and that $n_i > n_j$ when $i > j$, and that each element of g can be uniquely factorized into a product of the $a_i$s. We want to show that the order of $A_1a_2$ in $G/A_1$ is the same as the order of $a_2$ in $G$ and that this order is maximal. $\endgroup$ – Jacob FG May 17 '20 at 20:14
  • $\begingroup$ @JacobFG Since $a_i^{p^{n_i}}=e$ by assumption I can see that $(A_1a_2)^{p^{n_2}}=A_1(a_2^{p^{n_2}})=A_1e=A_1$, which means the order of $A_1a_2$ in $G/A_1$ is no greater than $p^{n_2}$. I think the next step is to show that, whenever $m,n$ are such that $1\leq m<n\leq p^{n_2}$, then $A_1a_2^m\neq A_1a_2^n$. Is this the right approach, and if so could you provide me with an additional hint? Thanks for your help! $\endgroup$ – buffle May 18 '20 at 2:43

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