# If ${\rm ord}(a) = n$ where $n$ is odd, then ${\rm ord}(a^2) = n$.

$$\DeclareMathOperator{\ord}{ord}$$ 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.

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 $$n | m$$?

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

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

• I follow you up to $n(k/2) = m$. How does that imply $n | m$? Commented Jul 21, 2019 at 0:29
• @dharmatech $\,k\,$ is even so $\,k/2\,$ is an integer. $\ \$ Commented Jul 21, 2019 at 0:32
• Ok... Thank you for all the help. So very appreciative of your patience. Commented Jul 21, 2019 at 0:34

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: $$\vert a^k\vert=\dfrac{\vert a\vert}{\operatorname {gcd}(\vert a\vert,k)}$$.

• 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.) Commented Jul 21, 2019 at 0:02
• It means $a$ and $b$ are relatively prime. I.e. $\operatorname {gcd}(a,b)=1$.
– user403337
Commented Jul 21, 2019 at 0:04
• OK. Thanks for your help Chris! Commented Jul 21, 2019 at 0:04
• You're welcome.
– user403337
Commented Jul 21, 2019 at 0:07

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, $${\rm ord}(a^2) = |\langle a^2\rangle| = |\langle a\rangle| = {\rm ord}(a)$$.