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Smallest numbers of element

Let d(\mathcal{G}) be the smallest number of elements necessary to generate a finite group \mathcal{G}. Show that |\mathcal{G}| \geq 2^{d (\mathcal{G}).

(by convention d(\mathcal{G})=0 if |\mathcal{G}|=1.)


We are using induction on d(\mathcal{G}). Clearly if d(\mathcal{G})=0 then |\mathcal{G}|=1. The result is also true if d(\mathcal{G})=1 since the non-identity element has order at least 2. Hence |\mathcal{G}| \geq 2. Let d(\mathcal{G} )=n. Assume that if a group H  is generated by n-1 elements, then |H| \geq 2^{n-1}. Let the generators of \mathcal{G} be \{x_1, \; x_2, \; \dots, \;  x_n \}. Then the subgroup

    \[\mathcal{T}=\langle x_1 , x_2, \dots, x_{n-1} \rangle\]

is a proper subgroup of \mathcal{G} and by assumption |\mathcal{T} | \geq 2^{n-1}. Since x_n \notin \mathcal{T} we obtain that x_n \mathcal{T} is a left coset of \mathcal{T} in \mathcal{G} and x_n \mathcal{T} \cap \mathcal{T} = 0. Moreover , x_n \mathcal{T} \cup T \subseteq \mathcal{G}. Hence

    \begin{align*} \left | \mathcal{G} \right | &\geq \left | x_n \mathcal{T} \cup \mathcal{T} \right | \\ &=\left | x_n \mathcal{T} \right | + \left | \mathcal{T} \right | \\ &=2 \left | \mathcal{T} \right | \\ &\geq 2 \cdot 2^{n-1} \\ &=2^n \end{align*}

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