Let \(p\) be a prime and let \(G\) be a group such that \(g^p=1\) for all \(g\in G\). Show that if \(p=2\) then \(G\) is abelian, and give an example with \(p>2\) for which \(G\) is not abelian.
Solution
1. If \(p=2\), then \(G\) is abelian.
For every \(g\in G\), \[ g^2=1, \] so \(g^{-1}=g\). Given \(g,h\in G\), the hypothesis also applies to \(gh\), hence \[ gh = (gh)^{-1} = h^{-1}g^{-1} = hg. \] Thus every two elements commute.
2. Let \(p>2\) be prime, and let \(U_3(\FF_p)\) be the subgroup of \(\GL_3(\FF_p)\) consisting of matrices\[ \begin{pmatrix} 1&a&b\\ 0&1&c\\ 0&0&1 \end{pmatrix}, \qquad a,b,c\in\FF_p. \]Every element of \(U_3(\FF_p)\) has \(p\)th power equal to the identity.
Every such matrix has the form \[ I+X, \] where \(X\) is strictly upper triangular. Hence \[ X^3=0. \] Since \(I\) and \(X\) commute, the binomial theorem gives \[ (I+X)^p = I+pX+\binom{p}{2}X^2, \] because all terms involving \(X^k\) with \(k\geq3\) vanish. In the field \(\FF_p\), \[ p=0 \] and, since \(p\) is odd, \[ \binom{p}{2} = \frac{p(p-1)}2 = 0. \] Therefore \[ (I+X)^p=I. \]
3. The group \(U_3(\FF_p)\) is not abelian.
Let \[ g = I+E_{12} = \begin{pmatrix} 1&1&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}, \qquad h = I+E_{23} = \begin{pmatrix} 1&0&0\\ 0&1&1\\ 0&0&1 \end{pmatrix}. \] Since \[ E_{12}E_{23}=E_{13} \qquad\text{and}\qquad E_{23}E_{12}=0, \] one has \[ gh = I+E_{12}+E_{23}+E_{13} \neq I+E_{12}+E_{23} = hg. \] Thus \(U_3(\FF_p)\) is nonabelian.
4. Hence exponent \(2\) forces a group to be abelian, whereas for every odd prime \(p\) there is a nonabelian group all of whose elements satisfy \(g^p=1\).
Step 1 proves the assertion for \(p=2\). Steps 2 and 3 give the required counterexample for every \(p>2\).
5. Q.E.D.
Step 4 is exactly the required conclusion.