After learning power laws and conjugation power formulas, we now apply them to group equations. These problems show how a small relation involving two elements can force a high power to become the identity. The key method is to rewrite conjugation expressions as powers, apply the same conjugation in two different ways, and then compare the results. In this lesson, students will solve representative power-equation problems and learn how to keep the order of factors correct in a non-commutative group. The main tool is the following observation. If $a^2=e$, then $a^{-1}=a$. Therefore an expression such as $a\circ b^r\circ a$ may be treated as $a\circ b^r\circ a^{-1}$, which is a conjugation. Conjugation respects powers: $$\begin{equation*} a\circ b^{rn}\circ a^{-1}=(a\circ b^r\circ a^{-1})^n. \end{equation*}$$ This allows one relation involving $b^r$ to produce relations involving higher powers of $b$. :::solved-problem Let $(G,\circ)$ be a group with identity element $e$ and let $a,b\in G$. If $$\begin{equation*} a^2=e \end{equation*}$$ and $$\begin{equation*} a\circ b^2\circ a=b^3, \end{equation*}$$ prove that $$\begin{equation*} b^5=e. \end{equation*}$$ ::: :::solution Let $(G,\circ)$ be a group with identity element $e$ and let $a,b\in G$. Given that $$\begin{equation*} a^2=e \end{equation*}$$ and $$\begin{equation*} a\circ b^2\circ a=b^3. \end{equation*}$$ To prove that $b^5=e$. Since $a^2=e$, we get $$\begin{equation*} a^{-1}=a. \end{equation*}$$ Therefore $$\begin{equation*} a\circ b^2\circ a^{-1}=b^3. \end{equation*}$$ Apply conjugation by $a$ to both sides. Since conjugation by $a$ is its own inverse when $a^2=e$, we get $$\begin{equation*} a\circ b^3\circ a^{-1}=b^2. \end{equation*}$$ Now compute $a\circ b^6\circ a^{-1}$ in two ways. First, $$\begin{align*} a\circ b^6\circ a^{-1} &=a\circ(b^2)^3\circ a^{-1}\\ &=(a\circ b^2\circ a^{-1})^3\\ &=(b^3)^3\\ &=b^9. \end{align*}$$ Second, $$\begin{align*} a\circ b^6\circ a^{-1} &=a\circ(b^3)^2\circ a^{-1}\\ &=(a\circ b^3\circ a^{-1})^2\\ &=(b^2)^2\\ &=b^4. \end{align*}$$ Therefore $$\begin{equation*} b^9=b^4. \end{equation*}$$ Multiplying both sides by $b^{-4}$, we get $$\begin{align*} b^9=b^4 &\implies b^9\circ b^{-4}=b^4\circ b^{-4}\\ &\implies b^5=e. \end{align*}$$ Hence, $$\begin{equation*} b^5=e. \end{equation*}$$ $\square$ ::: :::solved-problem Let $(G,\circ)$ be a group with identity element $e$ and let $a,b\in G$. If $$\begin{equation*} a^2=e \end{equation*}$$ and $$\begin{equation*} a\circ b^4\circ a=b^7, \end{equation*}$$ prove that $$\begin{equation*} b^{33}=e. \end{equation*}$$ ::: :::solution Let $(G,\circ)$ be a group with identity element $e$ and let $a,b\in G$. Given that $$\begin{equation*} a^2=e \end{equation*}$$ and $$\begin{equation*} a\circ b^4\circ a=b^7. \end{equation*}$$ To prove that $b^{33}=e$. Since $a^2=e$, we get $$\begin{equation*} a^{-1}=a. \end{equation*}$$ Therefore $$\begin{equation*} a\circ b^4\circ a^{-1}=b^7. \end{equation*}$$ Apply conjugation by $a$ to both sides. Since $a^{-1}=a$, conjugation by $a$ is its own inverse. Hence, $$\begin{equation*} a\circ b^7\circ a^{-1}=b^4. \end{equation*}$$ Now compute $a\circ b^{28}\circ a^{-1}$ in two ways. First, $$\begin{align*} a\circ b^{28}\circ a^{-1} &=a\circ(b^4)^7\circ a^{-1}\\ &=(a\circ b^4\circ a^{-1})^7\\ &=(b^7)^7\\ &=b^{49}. \end{align*}$$ Second, $$\begin{align*} a\circ b^{28}\circ a^{-1} &=a\circ(b^7)^4\circ a^{-1}\\ &=(a\circ b^7\circ a^{-1})^4\\ &=(b^4)^4\\ &=b^{16}. \end{align*}$$ Therefore $$\begin{equation*} b^{49}=b^{16}. \end{equation*}$$ Multiplying both sides by $b^{-16}$, we get $$\begin{align*} b^{49}=b^{16} &\implies b^{49}\circ b^{-16}=b^{16}\circ b^{-16}\\ &\implies b^{33}=e. \end{align*}$$ Hence, $$\begin{equation*} b^{33}=e. \end{equation*}$$ $\square$ ::: :::theorem Let $(G,\circ)$ be a group with identity element $e$ and let $a,b\in G$. If $$\begin{equation*} a^4=e \end{equation*}$$ and $$\begin{equation*} a^2\circ b=b\circ a, \end{equation*}$$ then $$\begin{equation*} a=e. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group with identity element $e$ and $a,b\in G$. Given that $$\begin{equation*} a^4=e \end{equation*}$$ and $$\begin{equation*} a^2\circ b=b\circ a. \end{equation*}$$ To prove that $a=e$. From $a^2\circ b=b\circ a$, multiply on the right by $b^{-1}$ to get $$\begin{equation*} a^2=b\circ a\circ b^{-1}. \end{equation*}$$ Now $$\begin{align*} b^2\circ a\circ b^{-2} &=b\circ(b\circ a\circ b^{-1})\circ b^{-1}\\ &=b\circ a^2\circ b^{-1}\\ &=(b\circ a\circ b^{-1})^2\\ &=(a^2)^2\\ &=a^4\\ &=e. \end{align*}$$ Multiplying $b^2\circ a\circ b^{-2}=e$ on the left by $b^{-2}$ and on the right by $b^2$, we get $$\begin{align*} b^2\circ a\circ b^{-2}=e &\implies a=b^{-2}\circ e\circ b^2\\ &\implies a=b^{-2}\circ b^2\\ &\implies a=e. \end{align*}$$ Hence, $$\begin{equation*} a=e. \end{equation*}$$ $\square$ ::: The last theorem illustrates why conjugation is useful. The relation $a^2\circ b=b\circ a$ is rewritten as $b\circ a\circ b^{-1}=a^2$. Conjugating again by $b$ produces $b^2\circ a\circ b^{-2}$, and the right-hand side becomes $(a^2)^2=a^4=e$. Since conjugation of an element is $e$ only when the original element is $e$, the conclusion $a=e$ follows. :::remark Power-equation problems require careful side control. Multiplying by an inverse on the wrong side may produce a false equation in a non-commutative group. In the theorem above, the equation $a^2\circ b=b\circ a$ gives $a^2=b\circ a\circ b^{-1}$ by multiplying on the right by $b^{-1}$. It does not give $a^2=b^{-1}\circ b\circ a$. ::: The first two solved problems compare two ways of conjugating the same power of $b$. We observe the common exponent $rs$ and the two resulting exponents $s^2$ and $r^2$. The difference between those two exponents becomes the power of $b$ forced to equal the identity. Try $r=2,s=3$ to recover $b^5=e$, and try $r=4,s=7$ to recover $b^{33}=e$. :::scientific-preview[Power Equation Exponent Comparison] @libraries: plotly ::: Use the calculator to check the arithmetic behind the exponent comparisons in the first two solved problems. Choose two positive integers $r$ and $s$. The calculator compares the two ways of computing the conjugate of $b^{rs}$ and returns the exponent difference $|s^2-r^2|$. This models why equations such as $b^9=b^4$ and $b^{49}=b^{16}$ force a power of $b$ to be the identity. :::calculator[Power Equation Exponent Test] ::: :::exercise [1] If $a^2=e$ and $a\circ b^3\circ a=b^5$, what is $a^{-1}$? [2] If $a^2=e$ and $a\circ b^3\circ a=b^5$, prove that $a\circ b^5\circ a=b^3$. [3] If $a^2=e$ and $a\circ b^3\circ a=b^5$, prove that $b^{16}=e$. [4] If $a^2=e$ and $a\circ b\circ a=b^{-1}$, prove that $a\circ b^n\circ a=b^{-n}$. [5] In the theorem above, explain why $b^2\circ a\circ b^{-2}=e$ implies $a=e$. ::: :::answer [1] $a^{-1}=a$. [2] Since conjugation by $a$ is its own inverse, applying it to $a\circ b^3\circ a=b^5$ gives $a\circ b^5\circ a=b^3$. [3] Compute $a\circ b^{15}\circ a$ as $(a\circ b^3\circ a)^5=b^{25}$ and also as $(a\circ b^5\circ a)^3=b^9$. Thus $b^{25}=b^9$, so $b^{16}=e$. [4] Since $a=a^{-1}$, use the conjugation power formula: $a\circ b^n\circ a=(a\circ b\circ a)^n=(b^{-1})^n=b^{-n}$. [5] Multiply on the left by $b^{-2}$ and on the right by $b^2$ to get $a=e$. ::: :::faq Q: Why does $a^2=e$ imply $a^{-1}=a$? A: Because $a\circ a=e$, so $a$ is its own inverse. Q: Why are conjugation formulas useful in power-equation problems? A: They convert conjugates of powers into powers of conjugates. Q: Can we multiply by inverses on either side? A: No. In a non-commutative group, the side of multiplication matters. Q: What is the main strategy in these problems? A: Compute the same conjugated power in two ways and compare the resulting powers. ::: :::call-to-action[Order of an Element] subtitle: Continue to order of an element, where powers equal to the identity become the central object of study. button: Next Section | /abstract-algebra/introduction-to-groups/order-of-an-element/finite-and-infinite-order button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/powers-of-elements-in-groups/conjugation-and-powers :::