Criterion for a Finite Abelian Group to be Simple
 Table of Contents

# Criterion for a Finite Abelian Group to be Simple

 Theorem 1: Let \$G\$ be a finite abelian group. Then \$G\$ is a simple group if and only if the order of \$G\$ is prime.
• Proof: Since \$G\$ is an abelian group, every subgroup of \$G\$ is normal. Thus, proving that \$G\$ is a simple group if and only if the order of \$G\$ is prime is equivalent to proving that \$G\$ has no proper nontrivial subgroups if and only if the order of \$G\$ is prime.
• \$\Rightarrow\$ Suppose that the order of \$G\$ is composite. Since \$G\$ is a finite abelian group, from the A Converse to Lagrange's Theorem for Finite Abelian Groups page we have that if \$a \mid |G|\$ then \$G\$ has a subgroup of order \$a\$. Since the order of \$G\$ is composite, there exists an \$a\$ with \$1 < a < |G|\$ such that \$a \mid |G|\$. Thus \$G\$ has a subgroup of order \$a\$, and since \$a \neq 1, |G|\$ we see that \$G\$ has a proper nontrivial subgroup that MUST be normal since \$G\$ is abelian. So \$G\$ is not a simple group.
• \$\Leftarrow\$ Suppose that the order of \$G\$ is prime. Then \$G\$ is isomorphic to \$\mathbb{Z}_p\$ by the result on the Cyclic Groups and their Isomorphisms page. By Lagrange's Theorem, the only subgroups of \$\mathbb{Z}_p\$ are the trivial group and the whole group. So \$G\$ has no proper nontrivial subgroups, i.e., \$G\$ has no proper nontrivial normal subgroups. So \$G\$ is simple.
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