The conjugation tests are powerful, but many normal subgroups are recognised faster through cosets. The most important quick test is the index two test: if a subgroup has exactly two left cosets, then it is automatically normal. This happens because the subgroup itself accounts for one coset, and everything outside it accounts for the other. In this lesson we also study two useful coset criteria and a square condition that forces normality. These results are especially helpful in examinations, where a full conjugation calculation may be longer than necessary. :::definition[Index Two Subgroup] Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. Then $H$ is called an index two subgroup of $G$ if $$\begin{equation*} [G:H]=2. \end{equation*}$$ ::: :::theorem Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If $[G:H]=2$, then $H$ is normal in $G$. ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $G$, and $[G:H]=2$. To prove that $H$ is normal in $G$. Let $x\in G$. If $x\in H$, then $$\begin{equation*} xH=H=Hx. \end{equation*}$$ If $x\notin H$, then $xH$ is the unique left coset different from $H$, and $Hx$ is the unique right coset different from $H$. Since the two left cosets are $H$ and $G-H$, and the two right cosets are $H$ and $G-H$, we get $$\begin{equation*} xH=G-H=Hx. \end{equation*}$$ Therefore $$\begin{equation*} xH=Hx \quad \forall x\in G. \end{equation*}$$ Hence, $H$ is normal in $G$. $\square$ ::: :::theorem Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If $[G:H]=2$, then $$\begin{equation*} a^2\in H \quad \forall a\in G. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $G$, and $[G:H]=2$. To prove that $a^2\in H$ for every $a\in G$. Let $a\in G$. If $a\in H$, then $a^2\in H$. If $a\notin H$, then $$\begin{equation*} aH=G-H. \end{equation*}$$ Since $a^{-1}\notin H$, we get $$\begin{equation*} a^{-1}\in G-H. \end{equation*}$$ Therefore $a^{-1}\in aH$. Thus there exists $h\in H$ such that $$\begin{equation*} a^{-1}=a\circ h. \end{equation*}$$ Hence $$\begin{align*} a^{-1}=a\circ h &\implies e=a^2\circ h\\ &\implies a^2=h^{-1}\in H. \end{align*}$$ Hence, $a^2\in H$ for every $a\in G$. $\square$ ::: :::theorem Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If $$\begin{equation*} x^2\in H \quad \forall x\in G, \end{equation*}$$ then $H$ is normal in $G$. ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $G$, and $x^2\in H$ for every $x\in G$. To prove that $H$ is normal in $G$. Let $x\in G$ and let $h\in H$. Since $x\in G$, $$\begin{equation*} x^2\in H. \end{equation*}$$ Since $x\circ h\in G$, $$\begin{equation*} (x\circ h)^2\in H. \end{equation*}$$ Since $x^{-1}\in G$, $$\begin{equation*} x^{-2}\in H. \end{equation*}$$ Now $$\begin{align*} x\circ h\circ x^{-1} &=x\circ h\circ x\circ x^{-2}\\ &=(x\circ h)^2\circ h^{-1}\circ x^{-2}. \end{align*}$$ Since $(x\circ h)^2\in H$, $h^{-1}\in H$, and $x^{-2}\in H$, we get $$\begin{equation*} x\circ h\circ x^{-1}\in H. \end{equation*}$$ Therefore $$\begin{equation*} xHx^{-1}\subseteq H \quad \forall x\in G. \end{equation*}$$ Hence, $H$ is normal in $G$. $\square$ ::: :::theorem Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If every left coset of $H$ is a right coset of $H$, then $H$ is normal in $G$. ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $G$, and every left coset of $H$ is a right coset of $H$. To prove that $H$ is normal in $G$. Let $a\in G$. Since $aH$ is a left coset of $H$, there exists $b\in G$ such that $$\begin{equation*} aH=Hb. \end{equation*}$$ Since $a\in aH$, we get $a\in Hb$. Also $a\in Ha$. Therefore the right cosets $Hb$ and $Ha$ have non-empty intersection. Therefore $$\begin{equation*} Hb=Ha. \end{equation*}$$ Thus $$\begin{equation*} aH=Ha. \end{equation*}$$ Therefore $$\begin{equation*} aH=Ha \quad \forall a\in G. \end{equation*}$$ Hence, $H$ is normal in $G$. $\square$ ::: :::theorem Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If the product of two left cosets of $H$ is always a left coset of $H$, then $H$ is normal in $G$. ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $G$, and the product of two left cosets of $H$ is always a left coset of $H$. To prove that $H$ is normal in $G$. Let $a\in G$. By the given condition, $$\begin{equation*} (a^{-1}H)(aH) \end{equation*}$$ is a left coset of $H$. Since $$\begin{equation*} e=a^{-1}\circ e\circ a\circ e\in (a^{-1}H)(aH), \end{equation*}$$ this left coset contains $e$. Therefore $$\begin{equation*} (a^{-1}H)(aH)=H. \end{equation*}$$ Let $h\in H$. Then $$\begin{equation*} a^{-1}\circ h\circ a\in (a^{-1}H)(aH). \end{equation*}$$ Thus $$\begin{equation*} a^{-1}\circ h\circ a\in H. \end{equation*}$$ Therefore $$\begin{equation*} a^{-1}Ha\subseteq H \quad \forall a\in G. \end{equation*}$$ Replacing $a$ by $a^{-1}$ gives $$\begin{equation*} aHa^{-1}\subseteq H \quad \forall a\in G. \end{equation*}$$ Hence, $H$ is normal in $G$. $\square$ ::: :::solved-problem Prove that $A_3$ is normal in $S_3$. ::: :::solution Let $G=S_3$ and $H=A_3$. Since $|S_3|=6$ and $|A_3|=3$, $$\begin{equation*} [S_3:A_3]=\frac{6}{3}=2. \end{equation*}$$ By the index two criterion, $$\begin{equation*} A_3\trianglelefteq S_3. \end{equation*}$$ ::: We observe why index two forces normality. Enter compatible orders for $G$ and $H$, and watch how the partition collapses into only two pieces. When the index is two, every element outside $H$ must lie in the same outside coset on both the left and the right. This leaves no room for a left coset and a right coset to differ. :::scientific-preview[Index Two Coset Explorer] ::: The proof uses uniqueness. Once $H$ has index two, every element outside $H$ must generate the same outside part whether we multiply on the left or on the right. We use the numerical index and the square condition as fast tests for normality. Enter the order of $G$ and $H$ to compute the index. Then mark whether the square condition $x^2\in H$ is known for every $x\in G$. The calculator separates two sufficient conditions: index two implies normality, and the square condition also implies normality. :::calculator[Index Two Normality Calculator] ::: :::exercise 1. Prove that every subgroup of index two is normal. 2. If $[G:H]=2$, prove that $a^2\in H$ for every $a\in G$. 3. Suppose $x^2\in H$ for every $x\in G$. Prove that $H$ is normal in $G$. 4. Give an example of an index two subgroup. ::: :::answer 1. There are only two left cosets and two right cosets. For $x\notin H$, both $xH$ and $Hx$ equal $G-H$, so $xH=Hx$. 2. If $a\in H$, then $a^2\in H$. If $a\notin H$, then $a^{-1}\in aH$, so $a^{-1}=ah$ for some $h\in H$, and hence $a^2=h^{-1}\in H$. 3. For $x\in G$ and $h\in H$, $xhx^{-1}=(xh)^2h^{-1}x^{-2}\in H$, so $xHx^{-1}\subseteq H$. 4. The subgroup $A_3$ has index two in $S_3$. ::: :::faq[Frequently Asked Questions] Q: Why does index two force normality? A: Outside $H$, there is only one possible left coset and only one possible right coset, so they must both be $G-H$. Q: Does $a^2\in H$ for all $a\in G$ imply $[G:H]=2$? A: Not necessarily. The square condition is sufficient for normality, not a complete characterisation of index two subgroups. Q: Why do coset products matter? A: Quotient groups require multiplication of cosets, and that multiplication behaves well only when the subgroup is normal. ::: :::call-to-action[Continue to Central and Cyclic Sources] subtitle: Study normal subgroups produced by centres and cyclic structures. button: Next Lesson | published/abstract-algebra/subgroups-and-normal-subgroups/normal-subgroups/central-and-cyclic-sources-of-normal-subgroups.md button-ghost: Previous Lesson | published/abstract-algebra/subgroups-and-normal-subgroups/normal-subgroups/definition-and-first-examples-of-normal-subgroups.md :::