I am asked if $\{n, n^{2}, n^{3}\}$ forms a group under multiplication modulo $m$ where $m = n + n^{2} + n^{3}.$
As an example we see that $\{2, 4, 8\}$ does form a group modulo $14,$ with identity $8,$ but am stuck starting the proof for the general case. Thanks in advance.
On
Though one can verify this result by brute force calculation, one gains much more insight by examining the ring-theoretic structure governing the result. First, we have factorizations
$\quad \smash[t]{(\color{#0a0}{\overbrace{n^3\!+\!n^2\!+\!n}^{\large m}})}(n\!-\!1)\, =\, n(n^3\!-1),\,$ and $\ G = \langle n\rangle = \{1,n,n^2\}$ is a subgroup of $\,\Bbb Z/(n^3\!-1)\phantom{I^{I^{I^I}}}$
Chinese rem (CRT) $\,\Rightarrow\, \Bbb Z/(n^3\!-1)\times \Bbb Z/n \,\cong\, \Bbb Z/n(n^3\!-1)\,\ $ by $\,\ (a,b)\to\, \color{#c00}{n^3} a + (1\!-\!n^3)\, b$
A ring hom is a group hom, therefore $\,(G,0)\to \color{#c00}{n^3}G \,$ remains a group in $\,R = \Bbb Z/n(n^3\!-1),\,$ namely $\ n^3 G = {n^3}\{1,n,n^2\} = \{n^3,n^4,n^5\} = \{n^3,n,n^2\},\:$ since $\,\ n^4 = n\,$ in $\ \Bbb Z/(n^4\!-n)$
And, again, the image of $\ n^3 G \subset R\,$ remains a group in $\,R/\color{#0a0}m \,\cong\,\Bbb Z/\color{#0a0}m\ \ $ QED
We work modulo $m=n+n^2+n^3$. Since $m|n^4-n$, we have that $n^4=n$. This implies* $n^3=1$, so $n^3$ is the identity. Also, when $x,y\in\{n,n^2,n^3\}=G$, we have $xy\in G$, because $xy=n^an^b=n^{a+b}=n^c$, where $c-1=a+b-1\mod 3$. Take for example $x=n^2$ and $y=n^3$. Then, $xy=n^5=n^4\cdot n=n\cdot n=n^2$. So in the exponent, we may subtract $3$ when it is at least $4$, making it the modulo $3$ remainder. (Note that we should take $n^3$ instead of $n^0$ here, since $n^3$ is already in our group.)
*this is not always true, but since we don't actually use that $n^3=1$ but only $n^4=n$, everything is fine.