Number of generators of a finite group

Posted: May 11, 2010 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: , ,

Problem. Prove that a group G with |G| \leq 2^n, \ n \in \mathbb{N}, can be generated by n elements.

Solution.  Let 1 \neq g_1 \in G and put G_1=\langle g_1 \rangle. Then |G_1| \geq 2. Now, if possible, choose g_2 \in G \setminus G_1 and put G_2=\langle g_1,g_2 \rangle. Then G_2 \supseteq G_1 \cup g_2G_1 and therefore

|G_2| \geq |G_1 \cup g_2G_1|=|G_1| + |g_2G_1|=2|G_1| \geq 2^2.

Again, if possible, choose g_3 \in G \setminus G_2 and put G_3=\langle g_1,g_2,g_3 \rangle. Then

|G_3| \supseteq |G_2 \cup g_3G_2|=2|G_2| \geq 2^3.

Continuing this way we will eventually find some r \in \mathbb{N} and G_r=\langle g_1, g_2, \cdots , g_r \rangle  with |G_r| \geq 2^r and it is no longer possible to choose any g \in G \setminus G_r. That means \langle g_1,g_2, \cdots , g_r \rangle = G_r=G. But then we’ll have 2^n \geq |G|=|G_r| \geq 2^r and so r \leq n. Thus G can be genrated by r \leq n elements. \Box

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