Let H be a group in which h²=Ifor all h is an element of H.

Prove: Suppose |H|<∞.Let {h₁,h₂,…..hn} be minimal set of generators for H. then http://www.mathhelpforum.com/math-he...2647988b-1.gif

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- Nov 8th 2009, 05:19 PMapple2009Prove: 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 http://www.mathhelpforum.com/math-he...2647988b-1.gif - Nov 8th 2009, 11:00 PMNonCommAlg
H is abelian, because for any $\displaystyle a,b \in H: \ 1=(ab)^2=abab,$ which gives us $\displaystyle ba=a^{-1}b^{-1}=ab.$ thus every element of H is in the form $\displaystyle h_1^{k_1} h_2^{k_2} \cdots \cdots h_n^{k_n},$ where $\displaystyle k_j \in \{ 0,1 \}.$

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 $\displaystyle \{h_1, \cdots , h_n \}$ being a__minimal__set of generators:

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