Subgroups of Finite Groups with Unique Order are Normal Subgroups

# Subgroups of Finite Groups with Unique Order are Normal Subgroups

Proposition 1: Let $G$ be a finite group. If $G$ has only one subgroup $H$ of order $n \mid |G|$ then $H$ is a normal subgroup of $G$. |

**Proof:**For each $g \in G$ let $\phi_g : H \to gHg^{-1}$ be defined for all $h \in H$ by:

\begin{align} \quad \phi_g(h) = ghg^{-1} \end{align}

- We claim that each $\phi_g$ is a bijection. Let $h_1, h_2 \in H$ and suppose that $\phi_g(h_1) = \phi_g(h_2)$. Then $gh_1g^{-1} = gh_2g^{-1}$. Multiplying this equation on the left by $g^{-1}$ and on the right by $g$ gives us that $h_1 = h_2$. So $\phi_g$ is injective. Furthermore, if $ghg^{-1} \in gHg^{-1}$ then $h \in H$ is such that $\phi_g(h) = ghg^{-1}$. So $\phi_g$ is surjective. Thus $\phi_g$ is a bijection.

- Since $G$ is a finite group and $H$ is a subgroup of $G$ we see that $|H| = n < \infty$. So, since $\phi_g : H \to gHg^{-1}$ is a bijection we have that $|H| = n = |gHg^{-1}|$.

- So $gHg^{-1}$ is a subgroup of $G$ of order $n$. Since $G$ has only one such subgroup, we must have that $gHg^{-1} = H$. And since this holds for all $g \in G$, we have that $H$ is a normal subgroup of $G$. $\blacksquare$