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!
Best Answer
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.