Normal subgroups are stable enough to be combined in several useful ways. Intersections of normal subgroups remain normal, and products of normal subgroups behave better than products of arbitrary subgroups. These results prepare the algebra needed for quotient groups and later for isomorphism theorems. In this lesson we work with $H\cap K$, $HK$, and $\langle H\cup K\rangle$, always checking whether the relevant set is actually a subgroup before discussing normality. A common mistake is to assume that $HK$ is automatically a subgroup for arbitrary subgroups $H$ and $K$; it is not automatic without an additional condition. :::definition[Product of Subgroups] Let $(G,\circ)$ be a group and let $H,K$ be subgroups of $G$. The product of $H$ and $K$ is the subset $$\begin{equation*} HK=\{h\circ k:h\in H,\ k\in K\}. \end{equation*}$$ ::: :::theorem Let $(G,\circ)$ be a group. Then the intersection of two normal subgroups of $G$ is normal in $G$. ::: :::proof Given that $(G,\circ)$ is a group. To prove that the intersection of two normal subgroups of $G$ is normal in $G$. Let $H$ and $K$ be normal subgroups of $G$. Since $H$ and $K$ are subgroups of $G$, $H\cap K$ is a subgroup of $G$. Let $x\in G$ and let $a\in H\cap K$. Then $a\in H$ and $a\in K$. Since $H$ is normal in $G$, $$\begin{equation*} x\circ a\circ x^{-1}\in H. \end{equation*}$$ Since $K$ is normal in $G$, $$\begin{equation*} x\circ a\circ x^{-1}\in K. \end{equation*}$$ Therefore $$\begin{equation*} x\circ a\circ x^{-1}\in H\cap K. \end{equation*}$$ Thus $$\begin{equation*} x(H\cap K)x^{-1}\subseteq H\cap K \quad \forall x\in G. \end{equation*}$$ Hence, $H\cap K$ is normal in $G$. $\square$ ::: :::corollary Let $(G,\circ)$ be a group. Then the intersection of any non-empty family of normal subgroups of $G$ is normal in $G$. ::: :::proof Given that $(G,\circ)$ is a group and $\{H_i:i\in I\}$ is a non-empty family of normal subgroups of $G$. To prove that $\bigcap_{i\in I}H_i$ is normal in $G$. Let $$\begin{equation*} N=\bigcap_{i\in I}H_i. \end{equation*}$$ The intersection of subgroups is a subgroup, so $N$ is a subgroup of $G$. Let $x\in G$ and let $a\in N$. Then $a\in H_i$ for every $i\in I$. Since each $H_i$ is normal in $G$, $$\begin{equation*} x\circ a\circ x^{-1}\in H_i \quad \forall i\in I. \end{equation*}$$ Therefore $$\begin{equation*} x\circ a\circ x^{-1}\in \bigcap_{i\in I}H_i=N. \end{equation*}$$ Hence, $N$ is normal in $G$. $\square$ ::: :::theorem Let $(G,\circ)$ be a group and let $H,K$ be subgroups of $G$. If $$\begin{equation*} HK=KH, \end{equation*}$$ then $HK$ is a subgroup of $G$. ::: :::proof Given that $(G,\circ)$ is a group, $H,K$ are subgroups of $G$, and $HK=KH$. To prove that $HK$ is a subgroup of $G$. Since $e\in H$ and $e\in K$, $$\begin{equation*} e=e\circ e\in HK. \end{equation*}$$ Therefore $HK$ is non-empty. Let $x,y\in HK$. Then there exist $h_1,h_2\in H$ and $k_1,k_2\in K$ such that $$\begin{equation*} x=h_1\circ k_1 \end{equation*}$$ and $$\begin{equation*} y=h_2\circ k_2. \end{equation*}$$ Now $$\begin{align*} x\circ y^{-1} &=(h_1\circ k_1)\circ(h_2\circ k_2)^{-1}\\ &=h_1\circ k_1\circ k_2^{-1}\circ h_2^{-1}. \end{align*}$$ Since $k_1\circ k_2^{-1}\in K$ and $h_2^{-1}\in H$, we get $$\begin{equation*} (k_1\circ k_2^{-1})\circ h_2^{-1}\in KH. \end{equation*}$$ Since $KH=HK$, there exist $h_3\in H$ and $k_3\in K$ such that $$\begin{equation*} (k_1\circ k_2^{-1})\circ h_2^{-1}=h_3\circ k_3. \end{equation*}$$ Therefore $$\begin{align*} x\circ y^{-1} &=h_1\circ h_3\circ k_3\\ &\in HK. \end{align*}$$ Hence, $HK$ is a subgroup of $G$. $\square$ ::: :::theorem Let $(G,\circ)$ be a group and let $H,K$ be subgroups of $G$. If $K$ is normal in $G$, then $HK=KH$ and $HK$ is a subgroup of $G$. ::: :::proof Given that $(G,\circ)$ is a group, $H,K$ are subgroups of $G$, and $K$ is normal in $G$. To prove that $HK=KH$ and $HK$ is a subgroup of $G$. Let $h\in H$ and $k\in K$. Since $K$ is normal in $G$, $$\begin{equation*} h\circ k\circ h^{-1}\in K. \end{equation*}$$ Therefore $$\begin{align*} h\circ k &=(h\circ k\circ h^{-1})\circ h\\ &\in KH. \end{align*}$$ Thus $HK\subseteq KH$. Again, since $K$ is normal in $G$, $$\begin{equation*} h^{-1}\circ k\circ h\in K. \end{equation*}$$ Therefore $$\begin{align*} k\circ h &=h\circ(h^{-1}\circ k\circ h)\\ &\in HK. \end{align*}$$ Thus $KH\subseteq HK$. Therefore $$\begin{equation*} HK=KH. \end{equation*}$$ By the subgroup product criterion, $HK$ is a subgroup of $G$. Hence, $HK=KH$ and $HK$ is a subgroup of $G$. $\square$ ::: :::theorem Let $(G,\circ)$ be a group and let $H,K$ be normal subgroups of $G$. Then $HK$ is normal in $G$. ::: :::proof Given that $(G,\circ)$ is a group and $H,K$ are normal subgroups of $G$. To prove that $HK$ is normal in $G$. Since $K$ is normal in $G$, we get $HK=KH$ and $HK$ is a subgroup of $G$. Let $x\in G$ and let $u\in HK$. Then there exist $h\in H$ and $k\in K$ such that $$\begin{equation*} u=h\circ k. \end{equation*}$$ Now $$\begin{align*} x\circ u\circ x^{-1} &=x\circ h\circ k\circ x^{-1}\\ &=(x\circ h\circ x^{-1})\circ(x\circ k\circ x^{-1}). \end{align*}$$ Since $H$ is normal in $G$, $$\begin{equation*} x\circ h\circ x^{-1}\in H. \end{equation*}$$ Since $K$ is normal in $G$, $$\begin{equation*} x\circ k\circ x^{-1}\in K. \end{equation*}$$ Therefore $$\begin{equation*} x\circ u\circ x^{-1}\in HK. \end{equation*}$$ Thus $$\begin{equation*} xHKx^{-1}\subseteq HK \quad \forall x\in G. \end{equation*}$$ Hence, $HK$ is normal in $G$. $\square$ ::: :::theorem Let $(G,\circ)$ be a group and let $H,K$ be subgroups of $G$. If $HK=KH$, then $$\begin{equation*} \langle H\cup K\rangle=HK. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H,K$ are subgroups of $G$, and $HK=KH$. To prove that $\langle H\cup K\rangle=HK$. Since $HK=KH$, the set $HK$ is a subgroup of $G$. Since $e\in K$, every $h\in H$ satisfies $$\begin{equation*} h=h\circ e\in HK. \end{equation*}$$ Therefore $H\subseteq HK$. Since $e\in H$, every $k\in K$ satisfies $$\begin{equation*} k=e\circ k\in HK. \end{equation*}$$ Therefore $K\subseteq HK$. Thus $$\begin{equation*} H\cup K\subseteq HK. \end{equation*}$$ Since $\langle H\cup K\rangle$ is the smallest subgroup of $G$ containing $H\cup K$, $$\begin{equation*} \langle H\cup K\rangle\subseteq HK. \end{equation*}$$ Also every subgroup of $G$ containing $H\cup K$ contains every product $h\circ k$ with $h\in H$ and $k\in K$. Therefore $$\begin{equation*} HK\subseteq \langle H\cup K\rangle. \end{equation*}$$ Hence, $\langle H\cup K\rangle=HK$. $\square$ ::: :::theorem Let $(G,\circ)$ be a group and let $H,K$ be normal subgroups of $G$. If $$\begin{equation*} H\cap K=\{e\}, \end{equation*}$$ then $$\begin{equation*} h\circ k=k\circ h \quad \forall h\in H,\ \forall k\in K. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H,K$ are normal subgroups of $G$, and $H\cap K=\{e\}$. To prove that $h\circ k=k\circ h$ for every $h\in H$ and every $k\in K$. Let $h\in H$ and $k\in K$. Let $$\begin{equation*} c=h\circ k\circ h^{-1}\circ k^{-1}. \end{equation*}$$ Since $K$ is normal in $G$, $$\begin{equation*} h\circ k\circ h^{-1}\in K. \end{equation*}$$ Therefore $$\begin{equation*} c=(h\circ k\circ h^{-1})\circ k^{-1}\in K. \end{equation*}$$ Since $H$ is normal in $G$, $$\begin{equation*} k\circ h^{-1}\circ k^{-1}\in H. \end{equation*}$$ Therefore $$\begin{equation*} c=h\circ(k\circ h^{-1}\circ k^{-1})\in H. \end{equation*}$$ Thus $c\in H\cap K$. Since $H\cap K=\{e\}$, we get $c=e$. Therefore $$\begin{align*} h\circ k\circ h^{-1}\circ k^{-1}=e &\implies h\circ k\circ h^{-1}=k\\ &\implies h\circ k=k\circ h. \end{align*}$$ Hence, $h\circ k=k\circ h$ for every $h\in H$ and every $k\in K$. $\square$ ::: :::solved-problem Let $G$ be a group and let $H,K\trianglelefteq G$. Prove that $H\cap K\trianglelefteq G$ and $HK\trianglelefteq G$. ::: :::solution Let $H,K\trianglelefteq G$. By the intersection theorem, $$\begin{equation*} H\cap K\trianglelefteq G. \end{equation*}$$ Since $H$ and $K$ are both normal in $G$, the product theorem gives $$\begin{equation*} HK\trianglelefteq G. \end{equation*}$$ ::: We observe which operations preserve normality. Mark whether $H$ and $K$ are normal in $G$, whether their product set commutes as $HK=KH$, and whether the intersection is trivial. The display separates three conclusions: intersection normality, product subgroup status, and elementwise commutation across $H$ and $K$. Try turning off normality for one subgroup and notice which conclusions disappear. :::scientific-preview[Normal Subgroup Operations Explorer] ::: We combine the counting formula for products with the normality results in this lesson. Enter finite orders for $H$, $K$, and their intersection to compute $|HK|$. Then mark whether $H$ and $K$ are normal to decide whether the product subgroup is automatically normal in $G$. The number $|HK|$ counts the product set, while the checkboxes control the structural conclusion. :::calculator[Normal Product Calculator] ::: :::exercise 1. Prove that the intersection of two normal subgroups is normal. 2. Give a sufficient condition for $HK$ to be a subgroup. 3. If $H,K\trianglelefteq G$, prove that $HK\trianglelefteq G$. 4. If $H,K\trianglelefteq G$ and $H\cap K=\{e\}$, prove that elements of $H$ commute with elements of $K$. ::: :::answer 1. If $a\in H\cap K$, then $xax^{-1}$ lies in both $H$ and $K$, so it lies in $H\cap K$. 2. A sufficient condition is $HK=KH$. In particular, this holds when $K\trianglelefteq G$. 3. Since $K\trianglelefteq G$, $HK=KH$ and $HK$ is a subgroup. Conjugating $hk$ by any $x\in G$ gives $(xhx^{-1})(xkx^{-1})\in HK$. 4. The commutator $hkh^{-1}k^{-1}$ lies in both $H$ and $K$, hence equals $e$, so $hk=kh$. ::: :::faq[Frequently Asked Questions] Q: Is the product $HK$ always a subgroup? A: No. It is a subgroup under additional hypotheses such as $HK=KH$. Q: Why does normality help products? A: Normality allows factors to be moved across subgroup products through conjugation. Q: Does $H\cap K=\{e\}$ imply all of $G$ is abelian? A: No. It only implies that elements of the two normal subgroups commute with each other. ::: :::call-to-action[Continue to Normality inside Related Subgroups] subtitle: Study how normality changes when the surrounding group changes. button: Next Lesson | published/abstract-algebra/subgroups-and-normal-subgroups/normal-subgroups/normality-inside-related-subgroups.md button-ghost: Previous Lesson | published/abstract-algebra/subgroups-and-normal-subgroups/normal-subgroups/central-and-cyclic-sources-of-normal-subgroups.md :::