# Examples of Normal Subgroups of a Group

Let $G$ be a group and let $H$ be a subgroup of $G$. We have already proven the following equivalences:

**1)**$H$ is a normal subgroup of $G$.

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

**3)**$N_G(H) = G$.

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

We will now look at some examples of normal subgroups of groups.

## Example 1

Let $\varphi : G \to G$ by the identity isomorphism defined for all $g \in G$ by $\varphi(g) = g$. Since $\varphi$ is an isomorphism we have that $\varphi$ is injective, and so $\ker (\varphi) = \{ 1 \}$, the trivial group. Let $H = \{ 1 \}$. Then by (4) we have that the trivial subgroup is always a normal subgroup of any group $G$.

## Example 2

Let $\varphi : G \to \{ 1 \}$ be the homomorphism from $G$ to the trivial group $\{ (1) \}$. Clearly $\ker (\varphi) = G$. Let $H = G$. Then by (4) we have that the whole group is always a normal subgroup of any group $G$.

## Example 3

Let $G$ be an abelian group. Let $H$ be any subgroup of $G$. Let $g \in G$. Let $\sigma_g : H \to H$ be defined for all $h \in H$ by $\sigma_g(h) = ghg^{-1}$. Since $G$ is abelian and since $g, h, g^{-1} \in G$ we have that $\sigma_g(h) = ghg^{-1} = gg^{-1}h = h$ for all $h \in H$. Thus $\sigma_g(H) = H$, i.e., $gHg^{-1} = H$. By (2) we have that $H$ is a normal subgroup of $G$.

Therefore EVERY subgroup $H$ of an abelian group $G$ is a normal subgroup of $G$.

## Example 4

Let $G$ be a group. Recall that $Z(G) = C_G(G)$ is the center of $G$ and is defined to be the set of all elements of $G$ that commute with every element of $G$. We have already proven that $Z(G)$ is an abelian subgroup of $G$.

Let $H$ be a subgroup of $Z(G)$. Then for all $g \in G$ and all $h \in H \subseteq Z(G)$ we have that $gh = hg$. So $gH = Hg$ for all $g \in G$. So by definition, $H$ is a normal subgroup of $G$.

Therefore EVERY subgroup of $Z(G)$ is a normal subgroup of $G$.

In particular, Example 3 becomes a consequence of Example 4 since $G$ is abelian if and only if $G = Z(G)$.