Timeline for Non-isomorphic groups with isomorphic nth powers (and similarly in other categories)
Current License: CC BY-SA 2.5
4 events
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| Apr 28, 2010 at 23:21 | comment | added | Vipul Naik | [Earlier version deleted]. I think that if $mx \equiv ny \pmod r$, with $m,n$ relatively prime, then $x$ is a $n^{th}$ multiple and $y$ is a $m^{th}$ multiple mod $r$. So this approach cannot work, at least directly. | |
| Apr 28, 2010 at 22:57 | comment | added | JBL | Well, Steven and I were tempted to use essentially the same idea again, but it doesn't work without at least some nontrivial modification: the linked question gives the existence of a group $B$ such that $B^m \cong B^n$ if and only if $m \equiv n \pmod{5}$, and for this $B$ we have $B^2 \not\cong B^3$ and $(B^2)^2 \cong (B^3)^3$, but of course we also have that $B^2 \cong (B^4)^3$ is a cube and $B^3 \cong (B^4)^2$ is a square. | |
| Apr 28, 2010 at 22:18 | comment | added | Vipul Naik | Thanks! That's really nice and seems obvious in hindsight. I shouldn't have missed that. Do you have any idea about the other question? | |
| Apr 28, 2010 at 22:01 | history | answered | Steven Sam | CC BY-SA 2.5 |