## Tuesday, 27 March 2012

### 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  (3-cycle, 5-cycle, 7-cycle, ... , $p_m$-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.