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Suppose that $x_1,x_2,··· ,x_k$ are elements in some group. Show that $|x_1x_2x_3···x_k| = |x_2x_3···x_kx_1|$. Note that the order might be infinite.

I have no idea how to prove this formally. I know that this should be the case because to be a group, one of the criteria is to be associative, thus it doesn't matter how you multiply them and in what order. however, I don't know how to formally show this on paper.

Shaun
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user20194358
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1 Answers1

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Hint: Conjugation by any group element is an automorphism of the group.

For more of a hint, click, hover over, or tap the box below.

Observe that $$\begin{align} c_{x_i}: G&\to G,\\ g&\mapsto x_i^{-1}gx_i\end{align}$$ is an isomorphism. Isomorphisms preserve order. Therefore, with $y=x_1x_2\dots x_n$, we have $$\begin{align}|x_1x_2\dots x_n|&=|y|\\ &=|c_{x_1}(y)|\\ &=|x_1^{-1}yx_1|\\ &=|x_1^{-1}(x_1x_2\dots x_n)x_1|\\ &=|(x_1^{-1}x_1)x_2\dots x_nx_1|\\ &=|e(x_2\dots x_nx_1)|\\ &=|x_2\dots x_nx_1|.\end{align}$$

Shaun
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    i dont think we learned about that yet. but ill do my research . thank you – user20194358 Feb 12 '23 at 20:49
  • No problem, @user20194358. The concept is simple; the main barrier is the terminology. Look up what each word means and then recall/learn what isomorphisms do to the orders of elements. – Shaun Feb 12 '23 at 21:13
  • You can just prove that in any group $g$ and $hgh^{-1}$ have the same order, it's not hard. – Justin Young Feb 12 '23 at 22:00
  • That's pretty much what I did, @JustinYoung; only, I did so with more of a view to understanding why it works. – Shaun Feb 12 '23 at 22:13
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    I meant my comment for @user20194358 who had reservations about the concepts introduced in your answer. – Justin Young Feb 13 '23 at 13:39
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    @Shaun Thank you for your response. it is very helpful and I understood it. However, it would be impossible for me to come up with a function like this on my own. did you come up with this function on your own for this specific problem? or were many problems solved with this function, and you just memorized it ? – user20194358 Feb 13 '23 at 23:09
  • You're welcome, @user20194358. The function $c_{x_i}$ is known as, "conjunction by $x_i$". Conjunction appears throughout abstract algebra. It arises naturally in group theory in particular; for example, see one of the many definitions of a normal subgroup of a group: a subgroup $N$ of a group $G$ such that for all $g\in G$, $g^{-1}Ng=N$. Have a look at inner automorphisms to get some idea of the fundamental nature of conjunction. I work with group theory a lot and so I'm used to it out of familiarity. If you study it enough, you see patterns. You'll get there . . . – Shaun Feb 13 '23 at 23:18