- #1
Kiwi1
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I am attempting to answer the attached question. I have completed parts 1-4 and am struggling with part 5.
5. Prove that if [tex]a^{l_0}b_1^{l_1}...b_n^{l_n}=e[/tex] then [tex]a^{l_0}=b_1^{l_1}=...=b_n^{l_n}=e[/tex]
If |a|>|b1|>|b2|>...>|bn| then I could raise both sides of [tex]a^{l_0}b_1^{l_1}...b_n^{l_n}=e[/tex] to the power of |b1| and would get [tex](a^{l_l})^{|b_1|}=e[/tex] from which I conclude: [tex](a^{l_0})=e[/tex] carrying on by induction gives me the required result.
But I don't think I can prove that the order of a is STRICTLY greater than the order of all b's?
5. Prove that if [tex]a^{l_0}b_1^{l_1}...b_n^{l_n}=e[/tex] then [tex]a^{l_0}=b_1^{l_1}=...=b_n^{l_n}=e[/tex]
If |a|>|b1|>|b2|>...>|bn| then I could raise both sides of [tex]a^{l_0}b_1^{l_1}...b_n^{l_n}=e[/tex] to the power of |b1| and would get [tex](a^{l_l})^{|b_1|}=e[/tex] from which I conclude: [tex](a^{l_0})=e[/tex] carrying on by induction gives me the required result.
But I don't think I can prove that the order of a is STRICTLY greater than the order of all b's?