After establishing integral power laws, we now study a power formula involving conjugation. In group theory, an expression of the form $a\circ b\circ a^{-1}$ is called a conjugate of $b$ by $a$. Conjugation appears throughout abstract algebra, especially in normal subgroups, group actions, symmetry groups, and matrix similarity. The main power formula says that the $n$-th power of a conjugate is the conjugate of the $n$-th power. In this lesson, students will prove this formula carefully using induction. :::definition[Conjugate Element] Let $(G,\circ)$ be a group and let $a,b\in G$. The element $$\begin{equation*} a\circ b\circ a^{-1} \end{equation*}$$ is called the $conjugate$ of $b$ by $a$. ::: Conjugation should be read as a transformation of $b$ by surrounding it with $a$ on the left and $a^{-1}$ on the right. In matrix groups, this resembles the expression $ABA^{-1}$, which is used in similarity of matrices. In abstract groups, conjugation is a way to compare elements within the same group. It preserves many algebraic properties, including powers. :::theorem Let $(G,\circ)$ be a group with identity element $e$ and let $a,b\in G$. Then $$\begin{equation*} (a\circ b\circ a^{-1})^n=a\circ b^n\circ a^{-1}\quad \forall\, n\in\mathbb{N}. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group with identity element $e$ and $a,b\in G$. To prove that $$\begin{equation*} (a\circ b\circ a^{-1})^n=a\circ b^n\circ a^{-1}\quad \forall\, n\in\mathbb{N}. \end{equation*}$$ We prove the result by mathematical induction on $n$. For $n=1$, $$\begin{align*} (a\circ b\circ a^{-1})^1 &=a\circ b\circ a^{-1}\\ &=a\circ b^1\circ a^{-1}. \end{align*}$$ Therefore the result is true for $n=1$. Assume that the result is true for $n=k$, that is, $$\begin{equation*} (a\circ b\circ a^{-1})^k=a\circ b^k\circ a^{-1}. \end{equation*}$$ To prove that the result is true for $n=k+1$. Now $$\begin{align*} (a\circ b\circ a^{-1})^{k+1} &=(a\circ b\circ a^{-1})^k\circ(a\circ b\circ a^{-1})\\ &=(a\circ b^k\circ a^{-1})\circ(a\circ b\circ a^{-1}) \quad [\because\, \text{induction hypothesis}]\\ &=a\circ b^k\circ(a^{-1}\circ a)\circ b\circ a^{-1} \quad [\because\, \circ \text{ is associative}]\\ &=a\circ b^k\circ e\circ b\circ a^{-1}\\ &=a\circ b^k\circ b\circ a^{-1}\\ &=a\circ b^{k+1}\circ a^{-1}. \end{align*}$$ Therefore the result is true for $n=k+1$. Hence, by mathematical induction, $$\begin{equation*} (a\circ b\circ a^{-1})^n=a\circ b^n\circ a^{-1}\quad \forall\, n\in\mathbb{N}. \end{equation*}$$ $\square$ ::: The cancellation in the middle of the induction step is the key: $$\begin{equation*} a^{-1}\circ a=e. \end{equation*}$$ When two conjugated factors are multiplied, the middle $a^{-1}$ from the first factor and $a$ from the second factor cancel. This leaves the powers of $b$ next to each other. That is why the expression becomes $a\circ b^{k+1}\circ a^{-1}$. :::corollary Let $(G,\circ)$ be a group with identity element $e$ and let $a,b\in G$. Then $$\begin{equation*} (a^{-1}\circ b\circ a)^n=a^{-1}\circ b^n\circ a\quad \forall\, n\in\mathbb{N}. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group with identity element $e$ and $a,b\in G$. To prove that $$\begin{equation*} (a^{-1}\circ b\circ a)^n=a^{-1}\circ b^n\circ a\quad \forall\, n\in\mathbb{N}. \end{equation*}$$ Apply the previous theorem with $a^{-1}$ in place of $a$. Since $(a^{-1})^{-1}=a$, we get $$\begin{align*} (a^{-1}\circ b\circ a)^n &=a^{-1}\circ b^n\circ(a^{-1})^{-1}\\ &=a^{-1}\circ b^n\circ a. \end{align*}$$ Hence, $$\begin{equation*} (a^{-1}\circ b\circ a)^n=a^{-1}\circ b^n\circ a. \end{equation*}$$ $\square$ ::: :::example Let $A$ and $B$ be invertible matrices of the same size. Then $$\begin{equation*} (ABA^{-1})^n=AB^nA^{-1}. \end{equation*}$$ For $n=2$, $$\begin{align*} (ABA^{-1})^2 &=(ABA^{-1})(ABA^{-1})\\ &=AB(A^{-1}A)BA^{-1}\\ &=ABIBA^{-1}\\ &=AB^2A^{-1}. \end{align*}$$ This is the matrix version of the group theorem. ::: :::solved-problem Let $(G,\circ)$ be a group with identity element $e$ and let $a,b\in G$. Prove that $$\begin{equation*} (a\circ b^2\circ a^{-1})^3=a\circ b^6\circ a^{-1}. \end{equation*}$$ ::: :::solution Let $(G,\circ)$ be a group with identity element $e$ and let $a,b\in G$. To prove that $$\begin{equation*} (a\circ b^2\circ a^{-1})^3=a\circ b^6\circ a^{-1}. \end{equation*}$$ Using the conjugation power formula, $$\begin{align*} (a\circ b^2\circ a^{-1})^3 &=a\circ(b^2)^3\circ a^{-1}\\ &=a\circ b^6\circ a^{-1}. \end{align*}$$ Hence, $$\begin{equation*} (a\circ b^2\circ a^{-1})^3=a\circ b^6\circ a^{-1}. \end{equation*}$$ $\square$ ::: :::solved-problem Let $(G,\circ)$ be a group with identity element $e$ and let $a,b\in G$. If $$\begin{equation*} a\circ b\circ a^{-1}=b^{-1}, \end{equation*}$$ prove that $$\begin{equation*} a\circ b^n\circ a^{-1}=b^{-n}\quad \forall\, n\in\mathbb{N}. \end{equation*}$$ ::: :::solution Let $(G,\circ)$ be a group with identity element $e$ and let $a,b\in G$. Given that $$\begin{equation*} a\circ b\circ a^{-1}=b^{-1}. \end{equation*}$$ To prove that $$\begin{equation*} a\circ b^n\circ a^{-1}=b^{-n}\quad \forall\, n\in\mathbb{N}. \end{equation*}$$ Using the conjugation power formula, $$\begin{align*} a\circ b^n\circ a^{-1} &=(a\circ b\circ a^{-1})^n\\ &=(b^{-1})^n\\ &=b^{-n}. \end{align*}$$ Hence, $$\begin{equation*} a\circ b^n\circ a^{-1}=b^{-n}\quad \forall\, n\in\mathbb{N}. \end{equation*}$$ $\square$ ::: :::remark The formula $(a\circ b\circ a^{-1})^n=a\circ b^n\circ a^{-1}$ is not the same as distributing powers over arbitrary products. The special structure matters: the middle inverse and the next left factor cancel in every repetition. Without this conjugation pattern, powers of products cannot usually be simplified in a non-commutative group. ::: A conjugated product repeats the pattern $aba^{-1}aba^{-1}\cdots$. We observe the middle cancellations that leave one $a$ at the far left, one $a^{-1}$ at the far right, and all copies of $b$ grouped together. Change the exponent and notice that the number of surviving $b$ factors becomes exactly $n$. This visual pattern explains why $(a\circ b\circ a^{-1})^n=a\circ b^n\circ a^{-1}$. :::scientific-preview[Conjugation Word Reduction] @libraries: plotly ::: Use the calculator to check the conjugation power identity for invertible diagonal matrices. The example uses ordinary matrix multiplication with diagonal matrices, so the arithmetic is easy to display. The purpose is to see the pattern $(ABA^{-1})^n=AB^nA^{-1}$ in a concrete setting. Since diagonal entries commute, the numbers simplify cleanly while the displayed formula still highlights the conjugation structure. :::calculator[Conjugation Power Test] ::: :::exercise [1] Define the conjugate of $b$ by $a$. [2] Prove that $(a\circ b\circ a^{-1})^2=a\circ b^2\circ a^{-1}$. [3] Prove that $(a^{-1}\circ b\circ a)^n=a^{-1}\circ b^n\circ a$. [4] If $a\circ b\circ a^{-1}=b^2$, find $a\circ b^3\circ a^{-1}$. [5] Explain why the conjugation power formula does not require the group to be abelian. ::: :::answer [1] The conjugate of $b$ by $a$ is $a\circ b\circ a^{-1}$. [2] $(a\circ b\circ a^{-1})^2=a\circ b\circ(a^{-1}\circ a)\circ b\circ a^{-1}=a\circ b^2\circ a^{-1}$. [3] Apply the conjugation power theorem with $a^{-1}$ in place of $a$. [4] Since $a\circ b^3\circ a^{-1}=(a\circ b\circ a^{-1})^3=(b^2)^3=b^6$. [5] The proof uses associativity and inverse cancellation, not commutativity. ::: :::faq Q: What is conjugation in a group? A: Conjugation is the expression $a\circ b\circ a^{-1}$. Q: What is the power of a conjugate? A: $(a\circ b\circ a^{-1})^n=a\circ b^n\circ a^{-1}$. Q: Does this formula hold in non-commutative groups? A: Yes. It is especially useful in non-commutative groups. Q: Why do the middle terms cancel? A: In repeated products, each $a^{-1}$ is followed by $a$, and $a^{-1}\circ a=e$. ::: :::call-to-action[Power Equations] subtitle: Continue to worked problems where power identities are used to prove equations such as powers equal to the identity. button: Next Lesson | /abstract-algebra/introduction-to-groups/powers-of-elements-in-groups/power-equations-in-groups button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/powers-of-elements-in-groups/integral-power-laws :::