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The following is an exercise from "Guide to Abstract Algebra", 1st edition, Carol Whitehead.

Let $G$ be a group and $x \in G$. Suppose $o(x)= n$ and $m$ is an integer such that $\gcd(m, n) = d$. Prove that $o(x^m) = n/d$.

Here $o(x)$ is the order of the element $x$, the smallest positive integer $r$ such that $x^r = e$, the identity element.

Question: I am unsure of my solution. Is it correct?

My Solution Attempt:

First we write what the order means,

$$o(x) = n \implies x^n = e$$

Then,

$$ (x^n)^k = (e)^k = e $$

where $k$ is a positive integer.

Now we want to re-arrange this in terms of $x^m$, so we can write $k=mq$.

$$e = (x^n)^{mq} = (x^m)^{nq}$$

Now, $nq$ is the order of element $(x^m)$ if $nq$ is the smallest positive integer. This is the case when $nq=n/d$, because $d$ is the largest integer that still allows $n/d$ to remain an integer.

Hence $o(x^m) = n/d$.

My Thoughts:

$e=(x^m)^{nq}$ is clearly true, but in seeking to minimise the exponent ${nq}$ I can't justify why we have to retain $n$. If it can be justified, then the next step of choosing $q=1/d$ is justifiable.

MathFail
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Penelope
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    $d$ is not necessarily the largest integer that still allows $\frac{n}{d}$ remain an integer. For example if $n=12,m=3$ then $d=gcd(12,3)=3$ but of course $6>d$ and $\frac{n}{6}=2$ an integer. To prove what you want, try to show that $o(x^{m})$ divides $\frac{n}{d}$ (you've practically already done that) and at the same time $\frac{n}{d}$ divides $o(x^{m})$ – Fotis May 07 '23 at 01:27
  • At the same time $o(x^{m})$ can't be strictly less than $\frac{n}{d}$ is what I wanted to write, for some reason it won't allow me to edit my previous comment – Fotis May 07 '23 at 01:49
  • I have made a second attempt at a solution here: https://math.stackexchange.com/questions/4694673/order-of-xm-given-the-order-of-group-element-x-is-n-where-gcdm-n-d - I couldn't delete this question sadly. – Penelope May 07 '23 at 22:18

1 Answers1

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Now, $nq$ is the order of element $(x^m)$ if $nq$ is the smallest positive integer. This is the case when $nq=n/d$, because $d$ is the largest integer that still allows $n/d$ to remain an integer.

This is false, if you let $nq=n/d$, it gives

$$q=\frac{1}d$$

But in general, $m, n$ are not co-primes, hence $d=\gcd(m, n)\ge 2$, but $q\in\mathbb{N}^+$, which means

$$q\ge1>\frac{1}d$$

Contradiction.

MathFail
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  • hi @MathFail is this the same point Fotis makes in the previous comment, that $d$ is not the largest integer that allows $n/d$ to remain an integer? – Penelope May 07 '23 at 12:15
  • The equation you wrote is false, $nq=n/d$ holds if and only if $q=d=1$, which is trivial. @Penelope – MathFail May 07 '23 at 15:08