Simple Groups

Recall that if $G$ is a group and $H$ is a subgroup of $G$ then the following statements are equivalent:

• 1) $H$ is a normal subgroup of $G$, i.e., $gH = Hg$ for all $g \in G$.

• 2) $gHg^{-1} \subseteq H$ for all $g \in G$.

• 3) $N_G(H) = G$.

• 4) There exists a group homomorphism $\varphi$ on $G$ such that $H = \ker (\varphi)$.

Consider the group $G = \mathbb{Z}/4\mathbb{Z}$. Let $H = \langle 2 \rangle$. Since $G$ is an abelian group, $H$ is a normal subgroup of $G$. So $G$ has a proper and nontrivial normal subgroup.

On the other hand, if we instead consider the group $G = \mathbb{Z}/5\mathbb{Z}$ we see by Lagrange's theorem that if $H$ is a subgroup of $G$ then $|H| \mid 5$. Since $5$ is prime, $|H| = 1$ or $|H| = 5$. So the only subgroups $H$ of $G$ are the trivial and whole groups. In other words, $G$ has no proper and nontrivial normal subgroups.

More generally, there are groups $G$ that have nontrivial proper subgroups - none of which are normal. Such groups are given a special name.

Definition: Let $G$ be a group. Then $G$ is said to be a Simple Group if $G$ has no proper nontrivial normal subgroups.

Example 1

As mentioned above, $\mathbb{Z}/5\mathbb{Z}$ is a simple group. In fact, the same argument above shos that $\mathbb{Z}/p\mathbb{Z}$ is a simple group for all primes $p$.

Example 2

As we will see on the Criterion for a Finite Abelian Group to be Normal page that if $G$ is a finite abelian group then $G$ is a simple group if and only if $|G| = p$.