# Generating direct powers

Before continuing, prove or disprove: If ${G}$ is a group, the direct power ${G^{\times n}}$ is never generated by fewer than ${n}$ elements.

Certainly this is the case if ${G}$ is abelian, as in this case ${G^{\times n}}$ has a quotient of the form ${\mathbf{F}_p^n}$, i.e., an ${n}$-dimensional vector space over ${\mathbf{F}_p}$. This is also the case if ${n\leq 2}$. Examples of small size are therefore a little hard to find.

Here is a nice example: Denote the ${m}$th prime by ${p_m}$, and let ${G = A_{p_m}}$. Then I claim that ${G^{\times(m-1)}}$ can be generated with ${3}$ elements. Indeed, take ${\alpha,\beta\in G^{\times(m-1)}}$ to be in each factor equal to some two generators ${\tilde{\alpha},\tilde{\beta}}$ of ${G}$, and take ${\gamma}$ to be an element of the form

$\displaystyle (\sigma_3, \sigma_5, \sigma_7, \dots, \sigma_{p_m}),$

where, for each ${p}$, ${\sigma_p}$ is any ${p}$-cycle. Then the ${(p_2\cdots p_j p_{j+2}\cdots p_m)}$th power of ${\gamma}$ is a ${p_{j+1}}$-cycle in the ${j}$th factor. With ${\alpha}$ and ${\beta}$ we can generate all the conjugates of this cycle, and these together generate the whole ${j}$th factor by simplicity.