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Math Help - Prove: Suppose |H|<∞. then

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    Prove: Suppose |H|<∞. then

    Let H be a group in which h=I for all h is an element of H.
    Prove: Suppose |H|<∞. Let {h₁,h₂,..hn} be minimal set of generators for H. then
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    Quote Originally Posted by apple2009 View Post
    Let H be a group in which h=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!
    Last edited by NonCommAlg; November 9th 2009 at 12:37 AM.
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