# Thread:

1. ## Prove: Suppose |H|<∞. then

Let H be a group in which h²=I for all h is an element of H.
Prove: Suppose |H|<∞. Let {h₁,h₂,…..hn} be minimal set of generators for H. then

2. Originally Posted by apple2009
Let H be a group in which h²=I for all h is an element of H.
Prove: Suppose |H|<∞. Let {h₁,h₂,…..hn} be minimal set of generators for H. then
H is abelian, because for any $a,b \in H: \ 1=(ab)^2=abab,$ which gives us $ba=a^{-1}b^{-1}=ab.$ thus every element of H is in the form $h_1^{k_1} h_2^{k_2} \cdots \cdots h_n^{k_n},$ where $k_j \in \{ 0,1 \}.$

to finish the proof, you only need to show that such a presentation for an element of H is unique. this is a result of the set $\{h_1, \cdots , h_n \}$ being a minimal set of generators:

if $h_1^{k_1} h_2^{k_2} \cdots \cdots h_n^{k_n}=h_1^{j_1} h_2^{j_2} \cdots \cdots h_n^{j_n}$ and say $j_r=0, \ k_r=1,$ then $h_r \in <\{h_i: \ i \neq r \}>$ and so $\{h_i: \ i \neq r \}$ would be a smaller set of generators for H. contradintion!
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