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.

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