For given prime number $p \neq 2$, construct a non-Abelian group with exponent $p$.

We know that for $p=2$ it's impossible.

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For $p>2$, take the set of upper-triangular $3\times 3$ matrices with elements from the field with $p$-elements and $1$s on the diagonal. This has exponent $p$ (why?), order $p^3$ (why?) and has centre of order $p$ (so is non-abelian). To check that the centre has order $p$, you should prove that a group of order $p^3$ has centre of order $p^3$ or of order $p$ and then show that the group is non-abelian. This group is called the Heisenberg group over the field $\mathbb{F}_p$

Note that for $p=2$, this group has exponent four. Look up the classification of Extra-Special $p$-groups for more details about these groups.

For example, take $p=3$. Then this group has the following presentation. $$\langle x,y,z; x^3 = y^3 = z^3 = 1, yz = zyx, xy = yx, xz = zx\rangle$$ The group has order $27$ and has exponent three. Clearly it is non-abelian (as otherwise $x=1$ and the group is simply the non-cyclic abelian group of order nine. Which it isn't.). For other primes $p>2$, change each of the three $3$s in the presentation to a $p$.

  • $\begingroup$ Thanks. Interesting construction! $\endgroup$ – jack Jul 28 '13 at 16:32
  • $\begingroup$ You're welcome - I read it in a book a long time ago! It is the smallest example, in the sense that it cannot happen for $p^2$ (why?) and by the classification of Extra-Special $p$-groups, this is the only example for $p^3$. $\endgroup$ – user1729 Jul 28 '13 at 16:35

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