- #1
Kiwi1
- 106
- 0
I am attempting to answer the attached question. I have completed parts 1-4 and am struggling with part 5.
5. Prove that if a^{l_0}b_1^{l_1}...b_n^{l_n}=e then a^{l_0}=b_1^{l_1}=...=b_n^{l_n}=e
If |a|>|b1|>|b2|>...>|bn| then I could raise both sides of a^{l_0}b_1^{l_1}...b_n^{l_n}=e to the power of |b1| and would get (a^{l_l})^{|b_1|}=e from which I conclude: (a^{l_0})=e 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 a^{l_0}b_1^{l_1}...b_n^{l_n}=e then a^{l_0}=b_1^{l_1}=...=b_n^{l_n}=e
If |a|>|b1|>|b2|>...>|bn| then I could raise both sides of a^{l_0}b_1^{l_1}...b_n^{l_n}=e to the power of |b1| and would get (a^{l_l})^{|b_1|}=e from which I conclude: (a^{l_0})=e 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?
