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$\mathbf{The \ Problem \ is}:$ Let $G:= \Bbb{Z}_9\times \Bbb{Z}_9\times \Bbb{Z}_9$ and $H:= \langle(6,6,6)\rangle.$ Find the invariant factors and elementary divisors of the quotient group $G/H.$

$\mathbf{My \ Approach}:$ Note, $H=\{(0,0,0),(3,3,3),(6,6,6)\}.$ Then $|G/H|=3^6/3=3^5.$ Now, I was trying to find the isomorphism type of $G/H$ using the Fundamental Theorem for finitely generated abelian groups. Clearly, $G/H$ is not a cyclic group, so $G/H\not\cong {\Bbb{Z}}_{3^5}.$ Let $A:={\Bbb{Z}}_{81}\times {\Bbb{Z}}_3$ and $B:={\Bbb{Z}}_{27}\times {\Bbb{Z}}_9.$ Now, I was trying to compare the number of order $3$ elements of $G/H$ with that of $A$ and $B.$ Firstly, the set of all order $3$ elements of $A$ is $\text{Ord}_3(A)=\{(p,q)\in A\mid p\in \{0,27,54\}, q\in \{0,1,2\}, (p,q)\neq (0,0)\}.$ Similarly, $\text{Ord}_3(B)=\{(p,q)\in B\mid p\in \{0,9,18\}, q\in \{0,3,6\}, (p,q)\neq (0,0)\}.$ As $\text{Ord}_3(G/H)=\{(a,b,c)\in G/H\mid a,b,c\in \{0,3,6\}, (a,b,c)\neq (0,0,0)\}\cup \{(x,x,x)\in G/H\mid x\in \{1,2,4,5,7,8\}\},$ thus $G/H\not\cong A$ and $G/H\not\cong B.$ Let, $C:= {{\Bbb{Z}}_3}^{\times 5}$ and here $\text{Ord}_3(C)$ has $3^5-1$ elements of order $3$ implying $G/H\not\cong C.$ For $D:={{\Bbb{Z}}_3}^{\times 3}\times {\Bbb{Z}}_9,$ $\text{Ord}_3(D)$ has $3^4-1$ elements which again implies $G/H\not\cong D.$ The last two remaining groups(upto isomorphism) are $E:= {\Bbb{Z}}_{27}\times {\Bbb{Z}}_3\times {\Bbb{Z}}_3$ and $F:= {\Bbb{Z}}_3\times {\Bbb{Z}}_9\times {\Bbb{Z}}_9.$ But calculating $\text{Ord}_3(E)$ and $\text{Ord}_3(F)$ both gives $26$ elements but then the list becomes exhausted. I think I am making some foolish mistake. Can you please provide some hints? Thanks in advance for that?

Rabi Kumar Chakraborty
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    The standard method of solving problems of this type is by computing the Smith Normal Form of an associated matrix over the integers, in this case the $4 \times 3$ matrix with rows $(9\ 0\ 0)$, $(0\ 9\ 0)$, $(0\ 0\ 9)$, $(6\ 6\ 6)$. The calculation is straightforward in this example, and results in a matrix with non-zero diagonal entries $3,9,9$. – Derek Holt Jun 14 '24 at 09:09

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Note that the order of every element of $G$ is a divisor of $9$, and hence the order of every element of $G/H$ is also a divisor of $9$. That rules out many of the possible isomorphism types for $G/H$ right away, and in particular is enough to finish off your approach.

Greg Martin
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