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For $p$ prime, show that every nontrivial element of the Heisenberg group is of order $p$ for $p\geq 3$.

The Heisenberg group looks like

$G= \left\{\left({\begin{array}{ccc} 1 & a & c \\ 0 & 1 & b \\ 0 & 0 & 1 \end{array}}\right): a,b,c \in \mathbb{Z}/p\mathbb{Z}\right\} $

I was thinking of using induction, and showed it holds true for $p=3$, but I'm not sure how to show it will hold true for $p=n$.

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    $\begingroup$ In such a case, reasoning by induction on $p$ is never possible. Think that in particular, in the induction step, you should jump from a prime number to the next prime number. $\endgroup$ – Jean Marie Oct 1 '17 at 4:36
  • $\begingroup$ Take a look to (en.wikipedia.org/wiki/…) $\endgroup$ – Jean Marie Oct 1 '17 at 5:33
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If $A\in G$, then $A$ can be written as $A=I+N$, with $N$ nilpotent ($N^3 = 0$). Then $$A^p = I + \binom{p}{1}N + \binom{p}{2}N^2$$ Since $p\geq 3$, the two binomial coefficients are divisible by $p$, hence $A^p = I$.

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    $\begingroup$ [+1] It is maybe important for the OP to know that the binomial expansion $(A+B)^p$ is allowed when $A$ and $B$ commute, which is the case here. $\endgroup$ – Jean Marie Oct 1 '17 at 4:33
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Actually, $p$ has to be a prime for this to be true. If you do it in $\mathbb Z/n$, or more generally in a ring with every element of order dividing $n$, then you obtain a group, I think, in which every element has order dividing $n$.

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  • $\begingroup$ Oops, I meant to have for p prime. I went back and edited it in. $\endgroup$ – frostyfeet Oct 1 '17 at 3:56

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