Can anyone give me a specific example of this:

Let $G=\langle a\rangle$ be a finite cyclic group of order n. If $m\in \mathbb{Z}$, then $\langle a^m\rangle =\langle a^d\rangle$, where $d=\gcd(m,n)$ and $a^m$ has order $\frac{n}{d}$.

I'm having a hard visualizing it. Thanks :D


Take cylic group of order 12, and write it like $\langle x \rangle = \{1, x, x^2, x^3, x^4, x^5, x^6, x^7, x^8, x^9, x^{10}, x^{11}\}$, so e.g. $x\cdot x^{11} = x^{12} = 1$.

Then $x^2$ generates only even powers, so there are $12/2 = 6$ elements in $\langle x^2\rangle$ $\left(=\{x^2, x^4, x^6, x^8, x^{10}, 1\}\right)$.

Similarly, $x^{10}$ generates the same subgroup, you get the same elements in different order: $\{x^{10}, x^8, x^6, x^4, x^2, 1\}$. It generates the same subgroup as $x^2$ because $\text{gcd}(10, 12) = 2$, by Euclidean algorithm, implies there exist integers $a$ and $b$ such that $10a+12b=2$. This means that, for some $a$, eventually, $x^{10a} = x^2$, so $\langle x^{10} \rangle = \langle x^{2} \rangle$.

This is actually summing remainders modulo $12$ but I wrote it as polynomials to be consistent with D&F notation.


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