Having learned how cosets partition a group and how indices measure the size of that partition, the next structural question is whether cosets can themselves be multiplied without ambiguity. This question leads naturally to normal subgroups. A subgroup may sit inside a group in many ways, but a normal subgroup sits symmetrically: its left cosets and right cosets coincide. Normal subgroups are the subgroups from which quotient groups are built, so they form one of the central bridges from elementary group theory to homomorphisms and isomorphism theorems. Students often think that every subgroup behaves well with cosets; the first purpose of this lesson is to separate ordinary subgroup behaviour from normal subgroup behaviour. :::definition[Normal Subgroup] Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. Then $H$ is called a normal subgroup of $G$ if $$\begin{equation*} xH=Hx \quad \forall x\in G. \end{equation*}$$ It is denoted by $$\begin{equation*} H\trianglelefteq G. \end{equation*}$$ ::: The condition $xH=Hx$ does not say that $xh=hx$ for each single element $h\in H$. It says that the whole left coset determined by $x$ is equal to the whole right coset determined by $x$. In an abelian group the two ideas collapse, but in a non-abelian group they are different. Normality is therefore not a statement about one product; it is a statement about how the subgroup is positioned inside the whole group. :::definition[Simple Group] Let $(G,\circ)$ be a group. Then $G$ is called a simple group if the only normal subgroups of $G$ are $$\begin{equation*} \{e\} \end{equation*}$$ and $$\begin{equation*} G. \end{equation*}$$ ::: :::example Let $(G,\circ)$ be a group with identity element $e$. Then $\{e\}$ and $G$ are normal subgroups of $G$. For every $x\in G$, $$\begin{align*} x\{e\}&=\{x\}\\ &=\{e\}x. \end{align*}$$ Also, $$\begin{align*} xG&=G\\ &=Gx. \end{align*}$$ Hence, $\{e\}\trianglelefteq G$ and $G\trianglelefteq G$. $\square$ ::: :::theorem Let $(G,\circ)$ be a group. If $G$ is abelian, then every subgroup of $G$ is normal in $G$. ::: :::proof Given that $(G,\circ)$ is a group and $G$ is abelian. To prove that every subgroup of $G$ is normal in $G$. Let $H$ be a subgroup of $G$. Let $x\in G$. For every $h\in H$, since $G$ is abelian, $$\begin{equation*} x\circ h=h\circ x. \end{equation*}$$ Therefore every element of $xH$ is an element of $Hx$, and every element of $Hx$ is an element of $xH$. Therefore $$\begin{equation*} xH=Hx \quad \forall x\in G. \end{equation*}$$ Hence, $H$ is normal in $G$. $\square$ ::: :::definition[Centre of a Group] Let $(G,\circ)$ be a group. The centre of $G$ is the subset $$\begin{equation*} Z(G)=\{z\in G:z\circ x=x\circ z \quad \forall x\in G\}. \end{equation*}$$ ::: :::theorem Let $(G,\circ)$ be a group. Then $Z(G)$ is normal in $G$. ::: :::proof Given that $(G,\circ)$ is a group. To prove that $Z(G)$ is normal in $G$. Let $x\in G$. For every $z\in Z(G)$, $$\begin{equation*} x\circ z=z\circ x. \end{equation*}$$ Therefore every element of $xZ(G)$ is an element of $Z(G)x$, and every element of $Z(G)x$ is an element of $xZ(G)$. Therefore $$\begin{equation*} xZ(G)=Z(G)x \quad \forall x\in G. \end{equation*}$$ Hence, $Z(G)$ is normal in $G$. $\square$ ::: :::example Let $G=S_3$ and let $$\begin{equation*} A_3=\{e,(123),(132)\}. \end{equation*}$$ The subgroup $A_3$ has index two in $S_3$, so it is a normal subgroup of $S_3$. By contrast, the subgroup $$\begin{equation*} H=\{e,(12)\} \end{equation*}$$ is not normal in $S_3$ because conjugation by $(123)$ sends $(12)$ to $(23)$, which is not in $H$. Thus $S_3$ gives the first useful warning: a subgroup of a non-abelian group need not be normal. $\square$ ::: We observe normality in $S_3$ by comparing one left coset with the corresponding right coset. Choose a subgroup and a group element, then compare $xH$ with $Hx$. The subgroup $A_3$ passes every comparison because its left and right cosets always match. The subgroup generated by $(12)$ fails for some choices of $x$, which shows why an ordinary subgroup of a non-abelian group need not be normal. :::scientific-preview[Coset Symmetry Explorer] ::: Normality is a global coset condition. Matching one pair of cosets is not enough, but one mismatch is enough to prove non-normality. :::solved-problem Let $(\mathbb{Z},+)$ be the additive group of integers. Prove that $n\mathbb{Z}$ is normal in $\mathbb{Z}$ for every integer $n$. ::: :::solution Let $n\in \mathbb{Z}$. Since $(\mathbb{Z},+)$ is abelian, every subgroup of $\mathbb{Z}$ is normal in $\mathbb{Z}$. Also $n\mathbb{Z}$ is a subgroup of $\mathbb{Z}$. Therefore $$\begin{equation*} n\mathbb{Z}\trianglelefteq \mathbb{Z}. \end{equation*}$$ ::: We can often prove normality without listing every coset. Enter the order of a finite group and a subgroup, then mark any structural information you know. The calculator recognises the automatic cases from this lesson and from the index two criterion. If none of those sufficient tests applies, the output is inconclusive rather than negative. :::calculator[Normality Quick Test Calculator] ::: :::exercise 1. Prove that every subgroup of an abelian group is normal. 2. Give two normal subgroups of an arbitrary group $G$. 3. Decide whether every subgroup of $S_3$ is normal. 4. Prove that $Z(G)$ is a subgroup of $G$ and then use the theorem above to state its normality. ::: :::answer 1. If $G$ is abelian and $H\leq G$, then $xh=hx$ for every $x\in G$ and $h\in H$, so $xH=Hx$ for every $x\in G$. 2. The subgroups $\{e\}$ and $G$ are always normal in $G$. 3. No. For example, $\{e,(12)\}$ is not normal in $S_3$. 4. The centre is closed under products and inverses because elements commuting with every element of $G$ remain commuting under products and inverses. Since $xZ(G)=Z(G)x$ for every $x\in G$, $Z(G)\trianglelefteq G$. ::: :::faq[Frequently Asked Questions] Q: Does $xH=Hx$ mean $xh=hx$ for every $h\in H$? A: No. It means the two cosets are equal as sets. Elementwise commutativity is stronger and is not required for normality. Q: Are all subgroups normal? A: No. Every subgroup of an abelian group is normal, but non-abelian groups often have subgroups that are not normal. Q: Why are normal subgroups important? A: Normal subgroups are exactly the subgroups that allow cosets to form quotient groups. ::: :::call-to-action[Continue to Index Two and Coset Criteria] subtitle: Learn fast coset tests for normality, including the index two criterion. button: Next Lesson | published/abstract-algebra/subgroups-and-normal-subgroups/normal-subgroups/index-two-and-coset-criteria-for-normality.md button-ghost: Review Double Cosets | published/abstract-algebra/subgroups-and-normal-subgroups/double-cosets/double-coset-definition.md :::