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I'm currently a self-learner on Abstract Algebra and Group theory. The book that I'm currently reading is A First Course In Abstract Algebra by John B. Fraleigh. So I know this question might truly be a basic question which every student would be taught in class and there is confusion about it. However, maybe it is me who is not careful in reading the book that raises in my mind this question. So given a binary operation $\ast $ on a set $S$, how can we compute:
$$a \ast b \ast c$$ Actually, on page 23 of the book, the author does mention this but I still can not get it. I don't know if we should compute this from left to right or reversely or we will just leave it there if $\ast$ is not associative. Moreover, if this binary operation is associated, is it true to define that: $$a \ast b \ast c = \left(a \ast b \right)\ast c$$ Then, after learning about Cyclic group there is a problem of which can be stated as below:

Let $a$ and $b$ be elements of a group $G$. Show that if $ab$ has finite order $n$, then $ba$ also has order $n$.

I know that we should show that $(ba)^n = e$ which can be driven from $(ab)^n = e$. Unfortunately, since $G$ might not be an Abelian group so what I had gone so far before looking for a solution is just manipulating $(ab)^n = e$ by using the fact that $(ab)^{-1} = b^{-1} a^{-1}$ since I don't know how to move further after splitting $(ab)^n = (ab)(ab)^{n-1}$. I also get confused whether $(ab)^n = (ab)(ab)^{n-1}$ or $(ab)^n = (ab)^{n-1} (ab)$. So I checked up for the solution and this is the way they manipulate it that I found: Show that if ab has finite order n, then ba also has order n. - Fraleigh p. 47 6.46. So I completely have no idea about this can be infered:

\begin{align}(\color{darkorange}{a}\color{darkcyan}{b})^n &= e \implies \color{darkorange}{a}\color{darkcyan}{b}(ab)^{n-1} = e \implies \color{darkcyan}{b}(ab)^{n-1}\color{darkorange}{a} = e \implies (ba)^n = e \end{align} I also tried to check another link in order to understand the solution: same question And the asker states that: $$(ab)^n=\underbrace{(ab)(ab)(ab)\cdots(ab)}_{n~\text{copies}}$$ $$=a\underbrace{(ba)(ba)(ba)\cdots(ba)}_{n-1~\text{copies}}b$$ $$=a(ba)^{n-1} b$$ It is here that the number of terms which we're dealing with is $n$ now. I don't have any idea how could he manipulates from the first line to the second line so I think that I need some helps on this. In summary, how can we compute binary operation of more than 3 elements? Pleas explain the problems that I state above and give some other examples if possible. Thank you so much.

Shaun
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Nguyễn Quân
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2 Answers2

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Write

$$(ab)^n = \underbrace{ababab \ldots ab}_{n {\text{ times}}}$$

Then

$$(ba)^n = \underbrace{bababab \ldots ba}_{n {\text{ times}}} = (\underbrace{bababab \ldots ba}_{n {\text{ times}}})bb^{-1} = b(\underbrace{ababab \ldots ab}_{n {\text{ times}}})b^{-1} = b(ab)^nb^{-1}.$$

If $(ab)^n = e$, then the line above gives

$$(ba)^n = b(ab)^nb^{-1} = beb^{-1} = e.$$

Suppose we were to replace $e$ with any element $z \in Z$ that commutes with any other element of $G$. Then what would we have.

Mike
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Conjugation by any element of the group is an automorphism, and thus preserves order. All that is left is to note that $ab$ and $ba$ are conjugates: $b(ab)b^{-1}=ba$.