Classify all abelian groups of order \(80\) up to isomorphism.
Solution
For a finite abelian group \(G\), write \(G_{(p)}\) for its \(p\)-primary subgroup.
1. If \(G\) is an abelian group of order \(80=2^4\cdot 5\), then\[ G\cong G_{(2)}\times G_{(5)}, \]where \(\abs{G_{(2)}}=16\) and \(G_{(5)}\cong \ZZ/5\ZZ\).
This is the primary decomposition from the fundamental theorem of finite abelian groups. The \(5\)-primary part has order \(5\), hence is cyclic.
2. The possibilities for the \(2\)-primary part are exactly\[ \ZZ/16\ZZ,\qquad \ZZ/8\ZZ\times\ZZ/2\ZZ,\qquad \ZZ/4\ZZ\times\ZZ/4\ZZ, \]\[ \ZZ/4\ZZ\times(\ZZ/2\ZZ)^2,\qquad (\ZZ/2\ZZ)^4. \]
By the fundamental theorem of finite abelian groups, the isomorphism types of abelian groups of order \(2^4\) correspond to the partitions of \(4\): \[ 4,\qquad 3+1,\qquad 2+2,\qquad 2+1+1,\qquad 1+1+1+1. \] These give precisely the five groups displayed in the claim.
3. Therefore the abelian groups of order \(80\), up to isomorphism, areexactly\[ \boxed{ \begin{gathered} \ZZ/16\ZZ\times\ZZ/5\ZZ,\\ \ZZ/8\ZZ\times\ZZ/2\ZZ\times\ZZ/5\ZZ,\\ \ZZ/4\ZZ\times\ZZ/4\ZZ\times\ZZ/5\ZZ,\\ \ZZ/4\ZZ\times(\ZZ/2\ZZ)^2\times\ZZ/5\ZZ,\\ (\ZZ/2\ZZ)^4\times\ZZ/5\ZZ. \end{gathered} } \]
Step 1 reduces the classification to the \(2\)-primary part together with the unique \(5\)-primary part, and step 2 lists every possible \(2\)-primary part. The uniqueness clause in the fundamental theorem shows that the five displayed groups are pairwise nonisomorphic.
4. Q.E.D.
Step 3 gives the complete classification.