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There is a theorem in a textbook that says:

Let $G=\left \langle a \right \rangle$ with $|G|=n$. Then $G=\left \langle a^{k}\right \rangle$ iff $\gcd(k,n)=1$.

I don't understand the difference between $\left \langle a \right \rangle$ and $\left \langle a^k \right \rangle$? Is it for stating specific order of generator $a$ and implying that the cyclic group $G$ can have the same order as $a$?

Michael Hardy
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6 Answers6

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Take an easy example, e.g. $n=3$ and $k=2$. Then $\langle a \rangle =\{e,a,a^2\}$ and $\langle a^2\rangle =\{a^2,a^4,a^6\}=\{a^2,a,e\}$. So both are equal, because $a^3=e$.

Dietrich Burde
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This example should clear your doubt.

Let $G=D_8$, the symmetries of a square.

Observe that $|D_8|=8=n$, since we have $4$ rotations and $4$ reflections.

Note that $\langle r\rangle=\{1,r,r^2,r^3\}$ and $\langle r^2\rangle =\{1,r^2\}$.

Clearly, $\langle r\rangle \neq \langle r^2\rangle$, since $\gcd(n,k) = \gcd(2,8) = 2\neq 1$

Michael Hardy
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Roby5
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$1$ spans $(\mathbb Z/6\mathbb Z,+)$ but not $1+1 = 2$ since adding $2$ multiple times we only get $\{2,4,0\}.$

A proof of the statement, maybe it is not exactly the one in your book.

$\langle a^k\rangle = G \Leftrightarrow (a^k)^d = a$ for some $d \Leftrightarrow a^{kd-1}=e \Leftrightarrow n \mid kd -1 \Leftrightarrow kd -nl = 1 \Leftrightarrow \gcd(k,n)=1$

Michael Hardy
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  • Please note proper MathJax usage, as in my edits to this answer. In particular, I changed $<a^k>$ to $\langle a^k\rangle$ and $gcd(k,n)$ to $\gcd(k,n)$ and $n|kd-1$ to $n\mid kd-1$ (with \mid) and put the ${\text{curly braces}}$ in ${2,4,0}$ inside rather than outside MathJax. Writing $\gcd$ not only prevents italicization but also results in proper spacing in things like $a\gcd(b,c). \qquad$ – Michael Hardy Feb 03 '18 at 22:49
  • @MichaelHardy Thanks for editing, I will be more careful. –  Feb 04 '18 at 04:10
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It's saying $a$ and $a^k$ generate the exact same subgroup. For example $G=(Z_5,+)$, then $1,2,3,4$ all generate $G$.

Akababa
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Take for example $\frac{\mathbb Z}{4\mathbb Z}=\langle 1\rangle$ but $\langle 2\cdot 1\rangle \cong \frac{\mathbb Z}{2\mathbb Z}$. Then in this case the order of $1$, so the cardinality of the group is $4,$ and $\gcd(4,2)\neq 1$ so they do not generate the same group. Observe that, with simple proposition, you can deduce all the subgroups of a cyclic groups.

Michael Hardy
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Acuo95
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We could rewrite the theorem like this.

If $a$ is an generator of $G$, then $a^k$ is also generator of $G$, when $\gcd(n,k)=1$. So, the point is what the generator of $G$ is. You could see that $a$ and $a^k$ may be different in general case.

The statement told that it is not really different if there are a condition.

In addition, $G$ is cyclic because it generates by one generator and we know that the order of cyclic group is same with the order of its generators.

Michael Hardy
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ChoMedit
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