Criterion for a Finite Abelian Group to be Simple
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.