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Suppose, for cyclic group $\mathbb Z_n$, that $x$ is the generator. Then if $x$ and $x^a$ have the same order, then $\gcd(a, n) = 1$. I do not understand how $x$ and $x^a$ having the same order implies $\gcd(a, n) = 1$. Please help

George Law
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tralaman
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  • Lots of good answers here; I hope they help. One way you might have tried to answer the question for yourself before asking here would have been to work out all the powers of all the elements in $\mathbb{Z}_n$ for, say, $n = 5,6,7,9,12$, so you have some primes, a power of a prime and some other composites. Familiarity with small examples will be useful for lots of these kinds of questions. – Ethan Bolker Nov 24 '16 at 14:46

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If $a$ and $n$ had a nontrivial common divisor, say $d$, then we could take $x^a$ to the power of $n/d$ and get the identity element, so $x^a$ would have order at most $n/d$ which is less than $n$.

anon
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Consider the list of elements $$x^a,\ x^{2a},\ \ldots,\ x^{(n-1)a}$$ None of these is the identity since the order of $x^a$ is $n$. Also they are all distinct, for if $x^{ia}=x^{ja}$ for some $1\le i\le j\le n-1$ then $x^{(j-i)a}=e$ $\implies$ $i=j$. Hence all these elements together with the identity $e$ form the whole group, and so one of them must be the (non-identity) element $x$, i.e. $\exists\,r$ such that $$x^{ra}\ =\ x$$ Hence $x^{ra-1}\ =\ e$, i.e. $n$ (the order of $x$) divides $ra-1$. Hence there exists $s$ such that $$ra+sn\ =\ 1$$ that is to say, $\gcd(a,n)=1$.

Olinguito
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George Law
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Ok lets say that $gcd(a,n)=b$ and $b>1$. Then $b|a$ and $b|n$. We also see that $(n/b)*a = kn$ for some integer $k$. But that implies $(x^a)^{n/b} = x^{kn}=1$ so $x^a$ has order $\leq n/b$.

hjr
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$\ \color{#c00}{D\mid A,N} \,\Rightarrow\, (X^{\large A})^{\Large \color{#c00}{\frac{N}D}} = (X^{\large N})^{\Large \color{#c00}{\frac{A}D}} = 1^{\Large \frac{A}D} = 1,\ $ so $\,X^{\large A}\,$ has order $\,\le { \dfrac{N}D}\,\ {\large[}< N\ $ if $\ D>1\,{\large]}$

Remark $\ $ Generally $\ {\rm ord}\,X = N\,\Rightarrow\, {\rm ord}(X^{\large A}) = N/\gcd(N,A).\ $ See here for proofs.

Community
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Bill Dubuque
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Intuitively, this is because if gcd(a,n) = 1, then $(a)(n) = j$ where $j$ is the lowest common multiple between them. If you investigate the order, you know that generally, $$x^a(x^a) = x^n(x^{2a-n}) = x^{2a-n}$$ (Note that 2 is assumed to be the smallest element such that $2a - n > 0$. You need to successively traverse the mod cycles of $a$ and $n$, until you reach that $$\underbrace{x^ax^ax^ax^ax^ax^ax^a....}_{\text{i times}} =x^{ia-n}$$ Such that $ia - n \equiv 0 \mod n$, And by definition of gcd, $i = n$.

q.Then
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