The class equation studies how individual elements move under conjugation. The same idea applies to subgroups: a subgroup can be moved to another subgroup by conjugating all its elements at once. Conjugate subgroups have the same internal group structure, although they may sit in different positions inside the larger group. This distinction is important in Sylow theory, where Sylow subgroups are often not equal but are conjugate, and where normality is detected by invariance under conjugation. :::definition[Conjugate Subgroup] Let $G$ be a group, let $H$ be a subgroup of $G$, and let $a \in G$. The conjugate of $H$ by $a$ is the set $$\begin{equation*} aHa^{-1}=\{aha^{-1}\mid h \in H\}. \end{equation*}$$ ::: :::theorem Let $G$ be a group, let $H$ be a subgroup of $G$, and let $a \in G$. Then $aHa^{-1}$ is a subgroup of $G$. ::: :::proof Given that $G$ is a group, $H$ is a subgroup of $G$, and $a \in G$. To prove that $aHa^{-1}$ is a subgroup of $G$. Since $e \in H$, we have $$\begin{equation*} aea^{-1}=e. \end{equation*}$$ Therefore $e \in aHa^{-1}$, so $aHa^{-1}$ is non-empty. Let $ah_1a^{-1},ah_2a^{-1} \in aHa^{-1}$, where $h_1,h_2 \in H$. Then $$\begin{align*} (ah_1a^{-1})(ah_2a^{-1})^{-1} &= (ah_1a^{-1})(ah_2^{-1}a^{-1}) \\ &= a(h_1h_2^{-1})a^{-1}. \end{align*}$$ Since $H$ is a subgroup of $G$, we have $h_1h_2^{-1} \in H$. Therefore $a(h_1h_2^{-1})a^{-1} \in aHa^{-1}$. Hence $aHa^{-1}$ is a subgroup of $G$. $\square$ ::: :::theorem Let $G$ be a group, let $H$ be a subgroup of $G$, and let $a \in G$. Then $H$ is isomorphic to $aHa^{-1}$. ::: :::proof Given that $G$ is a group, $H$ is a subgroup of $G$, and $a \in G$. To prove that $H$ is isomorphic to $aHa^{-1}$. Define $f:H\to aHa^{-1}$ by $$\begin{equation*} f(h)=aha^{-1}. \end{equation*}$$ Let $h_1,h_2 \in H$. Then $$\begin{align*} f(h_1h_2) &= a(h_1h_2)a^{-1} \\ &= (ah_1a^{-1})(ah_2a^{-1}) \\ &= f(h_1)f(h_2). \end{align*}$$ Therefore $f$ is a homomorphism. Suppose that $f(h_1)=f(h_2)$. Then $ah_1a^{-1}=ah_2a^{-1}$. Multiplying on the left by $a^{-1}$ and on the right by $a$ gives $h_1=h_2$. Therefore $f$ is one-one. Let $y \in aHa^{-1}$. Then $y=aha^{-1}$ for some $h \in H$, and so $y=f(h)$. Therefore $f$ is onto. Therefore $f$ is an isomorphism. Hence $H \cong aHa^{-1}$. $\square$ ::: :::definition[Invariance Under an Element] Let $G$ be a group, let $H$ be a subgroup of $G$, and let $a \in G$. The subgroup $H$ is called invariant under $a$ if $$\begin{equation*} aHa^{-1}=H. \end{equation*}$$ ::: :::theorem Let $G$ be a group and let $H$ be a subgroup of $G$. Then $H$ is normal in $G$ if and only if $H$ is invariant under every element of $G$. ::: :::proof Given that $G$ is a group and $H$ is a subgroup of $G$. To prove that $H$ is normal in $G$ if and only if $H$ is invariant under every element of $G$. By definition, $H$ is normal in $G$ if and only if $$\begin{equation*} gHg^{-1}=H \end{equation*}$$ for every $g \in G$. This is exactly the statement that $H$ is invariant under every element of $G$. Hence $H$ is normal in $G$ if and only if $H$ is invariant under every element of $G$. $\square$ ::: :::definition[Conjugates Induced by a Subgroup] Let $G$ be a group and let $H,K$ be subgroups of $G$. The conjugates of $H$ induced by $K$ are the subgroups $$\begin{equation*} kHk^{-1},\quad k \in K. \end{equation*}$$ ::: :::theorem Let $G$ be a group and let $H,K$ be subgroups of $G$. Then every subgroup of the form $kHk^{-1}$, where $k \in K$, is a conjugate subgroup of $H$ in $G$. ::: :::proof Given that $G$ is a group and $H,K$ are subgroups of $G$. To prove that every subgroup of the form $kHk^{-1}$, where $k \in K$, is a conjugate subgroup of $H$ in $G$. Let $k \in K$. Since $K$ is a subgroup of $G$, we have $k \in G$. By the definition of a conjugate subgroup, $kHk^{-1}$ is a conjugate of $H$ by the element $k$. Hence every subgroup of the form $kHk^{-1}$, where $k \in K$, is a conjugate subgroup of $H$ in $G$. $\square$ ::: :::example Let $G=S_3$ and let $H=\{e,(12)\}$. Conjugating $H$ by elements of $S_3$ produces the subgroups generated by the three transpositions: $$\begin{equation*} \{e,(12)\},\quad \{e,(13)\},\quad \{e,(23)\}. \end{equation*}$$ These subgroups are not all equal, but they are conjugate in $S_3$. Each has order $2$, and each is isomorphic to the cyclic group of order $2$. ::: This preview shows what happens when a whole subgroup is conjugated at once. Choose a subgroup of $S_3$ and see the distinct subgroups obtained from it by conjugation. The transposition subgroup moves among three different order two subgroups, while $A_3$ stays fixed under every conjugation. Use the output to distinguish being isomorphic from being equal. This prepares the Sylow idea that subgroups may be different but still conjugate. :::scientific-preview[Conjugate Subgroup Orbit Explorer] ::: The orbit of a subgroup under conjugation records all positions that subgroup can occupy inside the ambient group. If the orbit has only one subgroup, then every conjugation returns the same subgroup. That is exactly normality expressed as invariance. This calculator checks the normality criterion in the finite examples from this page. Select a subgroup of $S_3$. The output lists whether conjugation fixes the subgroup for each type of conjugating element. The result is not just a yes-or-no test; it explains whether the subgroup is invariant under every element, which is the definition of normality used in the theorem. :::calculator[Invariance Test Calculator] ::: A conjugate subgroup is always isomorphic to the original subgroup, but normality asks for equality after every conjugation. The calculator highlights this difference by showing which elements fix the subgroup and which elements move it. :::solved-problem Let $H$ be a subgroup of a group $G$. Prove that $H$ is normal in $G$ if and only if $H$ is a union of conjugacy classes of elements of $G$. ::: :::solution Let $H$ be a subgroup of a group $G$. Suppose first that $H$ is normal in $G$. Let $h \in H$ and let $g \in G$. Since $H$ is normal in $G$, we have $gHg^{-1}=H$. Therefore $ghg^{-1}\in H$. Hence $Cl_G(h)\subseteq H$ for every $h \in H$. Therefore $$\begin{equation*} H=\bigcup_{h\in H}Cl_G(h). \end{equation*}$$ Thus $H$ is a union of conjugacy classes of $G$. Conversely, suppose that $H$ is a union of conjugacy classes of $G$. Let $h \in H$ and $g \in G$. Since $H$ is a union of conjugacy classes and $h \in H$, we have $Cl_G(h)\subseteq H$. Therefore $ghg^{-1}\in H$. Hence $gHg^{-1}\subseteq H$ for every $g \in G$. Replacing $g$ by $g^{-1}$ gives $g^{-1}Hg\subseteq H$. Multiplying this containment on the left by $g$ and on the right by $g^{-1}$ gives $H\subseteq gHg^{-1}$. Therefore $gHg^{-1}=H$ for every $g \in G$. Hence $H$ is normal in $G$. $\square$ ::: :::solved-problem Let $G$ be an abelian group and let $H$ be a subgroup of $G$. Prove that every conjugate of $H$ is equal to $H$. ::: :::solution Let $G$ be an abelian group and let $H$ be a subgroup of $G$. For $g \in G$ and $h \in H$, commutativity gives $gh=hg$. Therefore $$\begin{align*} ghg^{-1} &= hgg^{-1} \\ &= h. \end{align*}$$ Thus $gHg^{-1}\subseteq H$ for every $g \in G$. For every $h \in H$, we also have $h=ghg^{-1}$ because $ghg^{-1}=h$. Therefore $H\subseteq gHg^{-1}$. Hence $gHg^{-1}=H$ for every $g \in G$. Thus every conjugate of $H$ is equal to $H$. $\square$ ::: :::exercise [1] Let $H$ be a subgroup of $G$ and let $a \in G$. Prove that $|aHa^{-1}|=|H|$ when $H$ is finite. [2] Let $H$ be a normal subgroup of $G$. Find all conjugates of $H$. [3] In $S_3$, let $H=\{e,(123),(132)\}$. Prove that $H$ is invariant under every element of $S_3$. ::: :::answer [1] The map $h \mapsto aha^{-1}$ is a bijection from $H$ onto $aHa^{-1}$, so the two finite sets have equal cardinality. [2] If $H$ is normal in $G$, then $gHg^{-1}=H$ for every $g \in G$. Thus the only conjugate of $H$ is $H$ itself. [3] The subgroup $H=A_3$ has index $2$ in $S_3$, so it is normal. Therefore it is invariant under every element of $S_3$. ::: :::faq[Frequently Asked Questions] Q: Are conjugate subgroups always equal? A: No. Conjugate subgroups are isomorphic, but they need not be equal. Q: What does invariance under an element mean? A: It means conjugating the subgroup by that element returns the same subgroup. Q: How is normality related to invariance? A: A subgroup is normal exactly when it is invariant under every element of the ambient group. ::: :::call-to-action[Next Step] subtitle: Use invariance to define normalizers and count how many conjugate subgroups arise. button: Study Normalizers | published/abstract-algebra/sylow-theorems/conjugacy-classes/normalizers-and-counting-conjugate-subgroups button-ghost: Review Conjugate Subgroups | published/abstract-algebra/sylow-theorems/conjugacy-classes/conjugate-subgroups-and-invariance :::