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Supposing I've already got handy a proof that every finite abelian group can be written in at least one way as the direct product of some cyclic groups of prime-power order, is the following correct and sufficient to prove uniqueness? Is there an easier way to do it- in particular, can one avoid appealing to finite group cancellation (considering the cancellation is for a special case, anyway) while still presenting a similarly styled proof?

Suppose $G$ is a finite abelian group with two representations as a direct product of groups of prime power order- $G\cong \mathbb Z_{p_1^{a_1}} \times \mathbb Z_{p_2^{a_2}} \times ... \times \mathbb Z_{p_n^{a_n}}$ and $G\cong \mathbb Z_{q_1^{b_1}} \times \mathbb Z_{q_2^{b_2}} \times ... \times \mathbb Z_{q_m^{b_m}}$ where $p_1,p_2,...,p_n$ and $q_1,q_2,...,q_m$ are all primes (not necessarily distinct) and all of $a_1,a_2,...,a_n$ and $b_1,b_2,...,b_m$ are positive integers.
First, since $|G|=p_1^{a_1}p_2^{a_2}...p_n^{a_n}=q_1^{b_1}q_2^{b_2}...q_m^{b_m}$ all the primes appearing in either the $p$ sequence or the $q$ sequence must appear in the other (by the F.T.A.).
Let $r$ be a given prime that divides $|G|$, and let $p_{r_1},p_{r_2},...,p_{r_k}$ and $q_{r_1},q_{r_2},...,q_{r_j}$ be all the primes from each set respectively that equal $r$.
It must be that $p_{r_1}^{a_{r_1}}p_{r_2}^{a_{r_2}}...p_{r_k}^{a_{r_k}}=q_{r_1}^{b_{r_1}}q_{r_2}^{b_{r_2}}...q_{r_j}^{b_{r_j}}$ as these both equal the highest power of $r$ that divides $|G|$.
We then note that in any finite abelian group we will have only one largest $r$-group as, if $a^r=e$ and $b^r=e$, then $(ab)^r=a^rb^r=ee=e$.
In $|G|$, by the first representation this largest $r$-group equals (by Lagrange's theorem) $$<0>\times<0>\times...\mathbb Z_{p_{r_1}^{a_{r_1}}} \times \mathbb Z_{p_{r_2}^{a_{r_2}}} \times ... \times \mathbb Z_{p_{r_k}^{a_{r_k}}}\times...\times <0>$$
By the second, it equals $$<0>\times<0>\times...\mathbb Z_{q_{r_1}^{b_{r_1}}} \times \mathbb Z_{q_{r_2}^{b_{r_2}}} \times ... \times \mathbb Z_{q_{r_j}^{b_{r_j}}}\times...\times <0>$$
Since they are both isomorphic to the single largest $r$-group of $G$, they must be isomorphic to each other. But the first is isomorphic to $\mathbb Z_{p_{r_1}^{a_{r_1}}} \times \mathbb Z_{p_{r_2}^{a_{r_2}}} \times ... \times \mathbb Z_{p_{r_k}^{a_{r_k}}}$ and the second to $\mathbb Z_{q_{r_1}^{b_{r_1}}}\times \mathbb Z_{q_{r_2}^{b_{r_2}}} \times ... \times \mathbb Z_{q_{r_j}^{b_{r_j}}}$ so we have: $$\mathbb Z_{p_{r_1}^{a_{r_1}}} \times \mathbb Z_{p_{r_2}^{a_{r_2}}} \times ... \times \mathbb Z_{p_{r_k}^{a_{r_k}}}\cong \mathbb Z_{q_{r_1}^{b_{r_1}}}\times \mathbb Z_{q_{r_2}^{b_{r_2}}} \times ... \times \mathbb Z_{q_{r_j}^{b_{r_j}}}\\ \mathbb Z_{r^{a_{r_1}}} \times \mathbb Z_{r^{a_{r_2}}} \times ... \times \mathbb Z_{r^{a_{r_k}}}\cong \mathbb Z_{r^{b_{r_1}}}\times \mathbb Z_{r^{b_{r_2}}} \times ... \times \mathbb Z_{r^{b_{r_j}}}$$
Then, if $r^{a_{max}}$ (the largest of the $r^{a_{r_s}}$'s) is $\neq r^{b_{max}}$, we get that the order of the element with the highest order in the L.H.S. expression is different from the order of the element with the highest order in the R.H.S. expression contradicting their isomorphism. So $r^{a_{max}}=r^{b_{max}}$, i.e., $\mathbb Z_{r^{a_{max}}}\cong\mathbb Z_{r^{b_{max}}}$.
By the cancellation of finite groups from direct products: $$\mathbb Z_{r^{a_{r_1}}} \times \mathbb Z_{r^{a_{r_2}}} \times...\times \require{cancel} \cancel{\mathbb Z_{r^{a_{max}}}}\times...\times \mathbb Z_{r^{a_{r_k}}}\cong \mathbb Z_{r^{b_{r_1}}}\times \mathbb Z_{r^{b_{r_2}}} \times...\times \require{cancel} \cancel{\mathbb Z_{r^{b_{max}}}}\times...\times \mathbb Z_{r^{b_{r_j}}}$$
We then repeat this procedure of weeding out the largest $r^{a_{r_s}}$'s and $r^{b_{r_t}}$'s and cancelling them out until none are left. Since these were all the terms equal to a power of $r$ (a prime dividing $|G|$), the $r$-component of $G$'s decomposition (i.e., the cyclic groups that have order $=$ a power of $r$) must be unique. Since $r$ was arbitrary, $G$'s entire decomposition is unique (up to isomorphism).

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  • $\begingroup$ on first very quick read through it looks okay; I would formally state the “repeat this procedure of weeding out” as an induction on the smaller of $k$ and $j$; I would also order them in increasing order of size to make the notation easier. That would easily let you cancel the two larger ones using your ideas. But you kind of need to show that the cancellation is valid: ideall you have a lemma somewhere that says that for finite abelian groups, if $A\times B \cong A\times C$, then $B\cong C$. Also: use \langle and \rangle, not < and >. $\endgroup$ – Arturo Magidin Mar 4 at 5:49
  • $\begingroup$ @ArturoMagidin Thank you for your comment. Yeah, I guess that part of the argument could be rephrased in inductive terms. As for the validity of cancellation, I have linked above in a couple of places to a proof (there's a hyperlink at the top of the question)- thanks for the formatting tip, $\endgroup$ – Cardioid_Ass_22 Mar 4 at 5:55
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    $\begingroup$ You can avoide appealing to cancellation by instead using dimension of vector spaces and an inductive argument. Once you reduce to $p$-parts, you can consider $G/pG$ to verify that the number of direct factors are the same, viewing it as a vector space over $\mathbb{F}_p$. By considering the subgroups $p^iG$ and $G/p^iG$, plus setting $a_1\leq\cdots\leq a_r$ and $b_1\leq\cdots \leq b_r$ and $a_1+\cdots+a_r = b_1+\cdots+b_r$ you can get enough information to deduce $a_i=b_i$ for all $i$; but it’s a lot more work than cancellation and induction. $\endgroup$ – Arturo Magidin Mar 4 at 6:23

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