Let G be a finite, Abelian group. If G is not cyclic, then there exist a subgroup of G isomorphic to C x C. Proposition 4.3.6 Serge Lang
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There exists only two groups of order $p^2$ up to isomorphism.
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A finite group with the property that all of its proper subgroups are abelian
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In an abelian group, the elements of finite order form a subgroup.
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Can a normal subgroup of a finite nonabelian group be nonabelian?
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Converse of Lagrange's theorem for abelian groups
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In a group $G$ where $\exists!$ nontrivial, proper subgroup, show that $G$ is cyclic and $\lvert G\rvert=p^2$, for $p$ prime.
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Finite subgroups of the multiplicative group of a field are cyclic
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An abelian group $G$ with ${\rm Aut}(G)$ non-abelian
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Prove or disprove: The automorphism group of a finite cyclic group must be cyclic.
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consider the group $G=\mathbb Q/\mathbb Z$. For $n>0$, is there a cyclic subgroup of order n
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Proving $\phi(G)$ is cyclic if $\phi$ : $G \to H$ is group homomorphism and $G$ is cyclic
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Isomorphic abelian groups which are not isomorphic R-modules
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Isomorphic quotient groups $\frac{G}{H} \cong \frac{G}{K}$ imply $H \cong K$?
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Problem on abelian group
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Finite Abelian groups with the same number of elements for all orders are isomorphic
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${\rm Aut}(G)$ is cyclic $\implies G$ is abelian
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If $G/Z(G)$ is cyclic, then $G$ is abelian