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$\DeclareMathOperator{\ord}{ord}$ Exercise 10.D.4 from Pinter says:

Let $a$ be any element of finite order of a group $G$. Prove the following: If $\ord(a) = n$ where $n$ is odd, then $\ord(a^2) = n$.

Here's an approach I took.

We'll be using the following from the same chapter:

10.D.2

The order of $a^k$ is a divisor (factor) of the order of $a$.

10.T5 (Theorem 5)

Suppose an element $a$ in a group has order $n$. Then $a^t = e$ iff $t$ is a multiple of $n$.

Let's begin. By 10.D.2

$$\ord(a^2) \mid n$$

Let $m = \ord(a^2)$.

$$m \mid n$$

$$(a^2)^m = e$$

$$a^{2m} = e$$

By 10.T5

$$n \mid 2m$$

Due the the following rules:

$$even * even = even$$ $$even * odd = even$$ $$odd * even = even$$ $$odd * odd = odd$$

$2m$ must be even.

But $n$ divides it so $n$ must be a factor of $m$.

Thus we have:

$$ m | n$$ $$ n | m$$

And so

$$m = n$$

Question 1: Is this an OK approach? I realize there are other approaches, but I wanted to explore one that only uses facts presented in or before this chapter in the book.

Question 2: This page also presents a similar solution.

He ends up with the following, like the above approach:

$$m | n$$ $$ n | 2m$$

At this point he states that:

$\gcd(n, 2) = 1$ (because n is odd)

Hence $n | m$

I understand the fact that $\gcd(n, 2) = 1$. But, how does he go from that fact to $n | m$?

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    $\begingroup$ You have been around for seven years and a half. Haven't you yet noticed that you are supposed to use MathJax here? $\endgroup$ Commented Jul 20, 2019 at 22:44
  • $\begingroup$ @JoséCarlosSantos There appears to be no explicit rule saying that questions must use LaTeX. $\endgroup$
    – dharmatech
    Commented Jul 24, 2019 at 0:31

3 Answers 3

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Question 1: Is this an OK approach?

The argument that $\, o(a^2) =: m\mid n\mid 2m\,$ is correct. But the inference $\,n\mid m\,$ is not properly justified. To use parity, note $\, n\mid 2m\,\Rightarrow\, nk = 2m\,$ is even hence $k$ is even, by $\,n\,$ odd. So cancelling $2$ yields $\, n(k/2) = m,\,$ so $\, n\mid m\,$ as claimed.

At this point he states that:

gcd(n, 2) = 1 (because n is odd). Hence n | m

I understand the fact that gcd(n, 2) = 1. But, how does he go from that fact to n | m?

Likely they apply Euclid's Lemma $\, \gcd(n,a)=1,\ n\mid am\,\Rightarrow\, n\mid m$ or it's generalization below.

Theorem $\, \ m\mid cx \iff\, \dfrac{m}{(m,c)}\ {\Large \mid}\ x.\ \ \,$ Proof $\,\ $ Let $\ d = (m,c).\ $ Then

we deduce $\, \ m\mid cx \overset{{\rm cancel}\ d\!\!}\iff\ \color{#c00}{\dfrac{m}d}\ {\Large \mid}\ \color{#c00}{\dfrac{c}d}\:x\!\!\overset{\rm(EL)\!}\iff\! \dfrac{m}d\ {\Large \mid}\ x\,\ $ by Euclid's Lemma (EL),

because: $\,\ (m,c) = d\ \Rightarrow\, \color{#c00}{\left(\dfrac{m}d,\,\dfrac{c}d\right)} = (m,c)/d = 1\ $ by the GCD Distributive Law

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  • $\begingroup$ I follow you up to $n(k/2) = m$. How does that imply $n | m$? $\endgroup$
    – dharmatech
    Commented Jul 21, 2019 at 0:29
  • $\begingroup$ @dharmatech $\,k\,$ is even so $\,k/2\,$ is an integer. $\ \ $ $\endgroup$ Commented Jul 21, 2019 at 0:32
  • $\begingroup$ Ok... Thank you for all the help. So very appreciative of your patience. $\endgroup$
    – dharmatech
    Commented Jul 21, 2019 at 0:34
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It is a generalization of Euclid's lemma that: $a\mid bc \land (a,b)=1\implies a\mid c$. (This is sometimes called Gauß's lemma.)

Or, use the fact from the theory of cyclic groups: $\vert a^k\vert=\dfrac{\vert a\vert}{\operatorname {gcd}(\vert a\vert,k)}$.

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  • $\begingroup$ In your notation, I know that a | bc means "a divides bc". What does the (a, b) = 1 part mean? (I looked up Euclid's Lemma and that article didn't clarify.) $\endgroup$
    – dharmatech
    Commented Jul 21, 2019 at 0:02
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    $\begingroup$ It means $a$ and $b$ are relatively prime. I.e. $\operatorname {gcd}(a,b)=1$. $\endgroup$
    – user403337
    Commented Jul 21, 2019 at 0:04
  • $\begingroup$ OK. Thanks for your help Chris! $\endgroup$
    – dharmatech
    Commented Jul 21, 2019 at 0:04
  • $\begingroup$ You're welcome. $\endgroup$
    – user403337
    Commented Jul 21, 2019 at 0:07
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Alternatively, write $n=2k+1$. Then $a=ea=a^n a=a^{n+1}=a^{2k+2}=(a^2)^{k+1}$. Therefore $ \langle a^2\rangle \subseteq \langle a\rangle \subseteq \langle a^2\rangle $ and so $ \langle a^2\rangle = \langle a\rangle $. Thus, ${\rm ord}(a^2) = |\langle a^2\rangle| = |\langle a\rangle| = {\rm ord}(a)$.

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