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May 26, 2014 at 22:23 answer added Derek Holt timeline score: 3
May 26, 2014 at 20:32 review First posts
May 26, 2014 at 20:36
May 26, 2014 at 20:22 comment added Anthony Quas This sounds very unlikely. If $G=G_1\times G_2$ where $G_2$ does not have this property, then you can choose generators that all have non-trivial parts in $G_1$ and $G_2$. But if $g$ is from the $G_2$ part, it will fail to have your property, right?
May 26, 2014 at 20:16 history asked user50402 CC BY-SA 3.0