Normality is always relative to an ambient group. A subgroup may be normal in one group but not in a larger group, and this is why the notation $H\trianglelefteq G$ must mention both $H$ and $G$. In this lesson we examine what happens when the surrounding group is replaced by an intermediate subgroup, by a product subgroup, or by a subgroup that intersects a normal subgroup. These results are often used silently later in quotient group arguments, so it is useful to prove them explicitly now. :::definition[Intermediate Subgroup] Let $(G,\circ)$ be a group and let $H,L$ be subgroups of $G$. The subgroup $L$ is called an intermediate subgroup between $H$ and $G$ if $$\begin{equation*} H\subseteq L\subseteq G. \end{equation*}$$ ::: :::theorem Let $(G,\circ)$ be a group and let $H,L$ be subgroups of $G$ such that $$\begin{equation*} H\subseteq L\subseteq G. \end{equation*}$$ If $H$ is normal in $G$, then $H$ is normal in $L$. ::: :::proof Given that $(G,\circ)$ is a group, $H,L$ are subgroups of $G$, $H\subseteq L\subseteq G$, and $H$ is normal in $G$. To prove that $H$ is normal in $L$. Since $H$ is normal in $G$, $$\begin{equation*} xHx^{-1}\subseteq H \quad \forall x\in G. \end{equation*}$$ Since $L\subseteq G$, this gives $$\begin{equation*} xHx^{-1}\subseteq H \quad \forall x\in L. \end{equation*}$$ Hence, $H$ is normal in $L$. $\square$ ::: :::remark The converse is not generally true. A subgroup can be normal in a smaller subgroup without being normal in the whole group. For example, every subgroup is normal in itself, but not every subgroup is normal in the larger group containing it. ::: :::theorem Let $(G,\circ)$ be a group and let $H,K$ be subgroups of $G$. If $K$ is normal in $G$, then $K$ is normal in $HK$. ::: :::proof Given that $(G,\circ)$ is a group, $H,K$ are subgroups of $G$, and $K$ is normal in $G$. To prove that $K$ is normal in $HK$. Since $K$ is normal in $G$, the product $HK$ is a subgroup of $G$. Also $$\begin{equation*} xKx^{-1}\subseteq K \quad \forall x\in G. \end{equation*}$$ Since $HK\subseteq G$, we get $$\begin{equation*} xKx^{-1}\subseteq K \quad \forall x\in HK. \end{equation*}$$ Hence, $K$ is normal in $HK$. $\square$ ::: :::theorem Let $(G,\circ)$ be a group and let $H,K$ be subgroups of $G$. If $K$ is normal in $G$, then $H\cap K$ is normal in $H$. ::: :::proof Given that $(G,\circ)$ is a group, $H,K$ are subgroups of $G$, and $K$ is normal in $G$. To prove that $H\cap K$ is normal in $H$. Since $H\cap K$ is a subgroup of $H$, it is enough to prove $$\begin{equation*} h(H\cap K)h^{-1}\subseteq H\cap K \quad \forall h\in H. \end{equation*}$$ Let $h\in H$ and let $x\in H\cap K$. Then $x\in H$ and $x\in K$. Since $h,x\in H$, we get $$\begin{equation*} h\circ x\circ h^{-1}\in H. \end{equation*}$$ Since $K$ is normal in $G$ and $h\in G$, we get $$\begin{equation*} h\circ x\circ h^{-1}\in K. \end{equation*}$$ Therefore $$\begin{equation*} h\circ x\circ h^{-1}\in H\cap K. \end{equation*}$$ Thus $$\begin{equation*} h(H\cap K)h^{-1}\subseteq H\cap K \quad \forall h\in H. \end{equation*}$$ Hence, $H\cap K$ is normal in $H$. $\square$ ::: :::example Let $G=S_3$, let $K=A_3$, and let $H=\langle(12)\rangle$. Since $A_3\trianglelefteq S_3$, the theorem gives $$\begin{equation*} H\cap K\trianglelefteq H. \end{equation*}$$ Here $$\begin{equation*} H\cap K=\{e\}, \end{equation*}$$ which is normal in $H$. This example is simple, but it captures the general principle: intersecting a subgroup with a normal subgroup produces a normal subgroup inside the smaller subgroup. $\square$ ::: :::solved-problem Let $H\subseteq L\subseteq G$ be subgroups. Suppose $H\trianglelefteq G$. Prove that $H\trianglelefteq L$ using cosets. ::: :::solution Let $x\in L$. Since $L\subseteq G$, we have $x\in G$. Since $H\trianglelefteq G$, $$\begin{equation*} xH=Hx. \end{equation*}$$ Therefore $$\begin{equation*} xH=Hx \quad \forall x\in L. \end{equation*}$$ Thus $H$ is normal in $L$. ::: :::solved-problem Let $K\trianglelefteq G$ and let $H\leq G$. Prove that $HK$ is a subgroup and that $K\trianglelefteq HK$. ::: :::solution Since $K\trianglelefteq G$, we get $$\begin{equation*} HK=KH. \end{equation*}$$ Therefore $HK$ is a subgroup of $G$. Now let $x\in HK$. Since $HK\subseteq G$ and $K\trianglelefteq G$, $$\begin{equation*} xKx^{-1}\subseteq K. \end{equation*}$$ Hence $$\begin{equation*} K\trianglelefteq HK. \end{equation*}$$ ::: We observe that normality can be inherited when the ambient group gets smaller, but not usually when it gets larger. Select which facts are known, and the diagram shows the guaranteed conclusions. If $H\trianglelefteq G$ and $H\subseteq L\subseteq G$, then $H\trianglelefteq L$ follows immediately. If only $H\trianglelefteq L$ is known, no conclusion about $H\trianglelefteq G$ is forced. :::scientific-preview[Ambient Normality Explorer] ::: We can combine the related-subgroup theorems with ordinary counting. Enter the orders of $H$, $K$, and $H\cap K$ to compute the size of $HK$. The normality checkbox records whether $K\trianglelefteq G$ is known, which is the hypothesis that makes $K\trianglelefteq HK$ and $H\cap K\trianglelefteq H$ automatic. The calculation and the normality conclusion are related but not the same kind of statement. :::calculator[Related Normality Calculator] ::: :::exercise 1. If $H\trianglelefteq G$ and $H\subseteq L\subseteq G$, prove that $H\trianglelefteq L$. 2. If $K\trianglelefteq G$ and $H\leq G$, prove that $K\trianglelefteq HK$. 3. If $K\trianglelefteq G$ and $H\leq G$, prove that $H\cap K\trianglelefteq H$. 4. Give an example showing that normality depends on the ambient group. ::: :::answer 1. For every $x\in L$, also $x\in G$, so $xHx^{-1}\subseteq H$. 2. Since $HK\subseteq G$, the conjugation inclusion $xKx^{-1}\subseteq K$ holds for every $x\in HK$. 3. For $h\in H$ and $x\in H\cap K$, the element $hxh^{-1}$ lies in $H$ because $H$ is a subgroup and lies in $K$ because $K\trianglelefteq G$. 4. The subgroup $\langle(12)\rangle$ is normal in itself but is not normal in $S_3$. ::: :::faq[Frequently Asked Questions] Q: If $H$ is normal in $L$ and $L$ is normal in $G$, must $H$ be normal in $G$? A: No. Normality is not generally transitive. Q: Why is $H\cap K$ normal in $H$ when $K$ is normal in $G$? A: Conjugating by elements of $H$ keeps the result inside $H$, and normality of $K$ keeps it inside $K$. Q: Why does $K\trianglelefteq G$ imply $K\trianglelefteq HK$? A: The elements of $HK$ are still elements of $G$, so the conjugation condition for $G$ applies to them. ::: :::call-to-action[Continue to Normalisers of Subgroups] subtitle: Study the largest subgroup in which a given subgroup is normal. button: Next Lesson | published/abstract-algebra/subgroups-and-normal-subgroups/normal-subgroups/normalisers-of-subgroups.md button-ghost: Previous Lesson | published/abstract-algebra/subgroups-and-normal-subgroups/normal-subgroups/products-and-intersections-of-normal-subgroups.md :::