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

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

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

• Since you're new, I'd like to give you some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are. That way, people won't tell you stuff you already know, and they can write answers at an appropriate level; also, people tend to be more willing to help if you show that you've tried the problem yourself. – Zev Chonoles Jul 28 '13 at 16:08

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 $p$.
• 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 $p^3$. – user1729 Jul 28 '13 at 16:35