Criteria for a Subgroup to be Normal

# Criteria for a Subgroup to be Normal

Recall from the Normal Subgroups page that if $(G, \cdot)$ is a group and $(H, \cdot)$ is a subgroup then $(H, \cdot)$ is said to be a normal subgroup of $G$ if $gH = Hg$ for all $g \in G$, that is, the left and right cosets of $H$ with representative $g$ are equal for all $g \in G$.

We will now look at some criteria for when a subgroup of a group will be normal.

 Theorem 1: Let $G$ be a group and let $H$ be a subgroup of $G$. The following statements are equivalent: a) $H$ is a normal subgroup of $G$. b) For all $g \in G$ we have that $gHg^{-1} \subseteq H$. c) For all $g \in G$ we have that $gHg^{-1} = H$, i.e., $N_G(H) = \{ g \in G : gHg^{-1} = H\} = G$.
• Proof of $a) \Rightarrow c)$ Suppose that $H$ is a normal subgroup of $G$. Then $gH = Hg$ for all $g \in G$. Fix $g \in G$. Let $z \in gH$. Then $z = gh'$ for some $h \in H$. Since $gH = Hg$ we have that $z \in Hg$ and so $z = h''g$ for some $h'' \in H$. Thus $gh' = h''g$. So $gh'g^{-1} = h''$. Since $h'$ and $h''$ are arbitrary elements of $H$, we have that $gHg^{-1} = H$.
• Let $h' \in H$. Then $h'g \in Hg$. Since $gH = Hg$ we have that $h'g \in gH$. So there exists an $h'' \in H$ such that $h'g = gh''$. So$h' = gh''g^{-1} \in gHg^{-1}$ which shows that $H \subseteq gHg^{-1}$.
• Now let $h' \in gHg^{-1}$. Then there exists an $h'' \in H$ such that $h' = gh''g^{-1}$. So $h'g = gh'' \in gH$. Since $gH = Hg$ we have that $h'g \in Hg$. So $h' \in H$. This shows that $H \supseteq gHg^{-1}$.
• Thus, for all $g \in G$ we have that $gHg^{-1} = H$.
• Proof of $c) \Rightarrow b)$ Trivial.
• Proof of $b) \Rightarrow a)$ Suppose that for all $g \in G$ we have that $gHg^{-1} \subseteq H$. Then $gH \subseteq Hg$. Now since $g \in G$ we have that $g^{-1} \in G$ and by hypothesis, $g^{-1}Hg \subseteq H$. Therefore $Hg \subseteq gH$. So $gH = Hg$ for all $g \in G$, i.e., $H$ is a normal subgroup of $G$. $\blacksquare$