MHB Basis Theorem for Finite Abelian Groups

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 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?
 

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Hi,
What is needed is exercise O, which I have included in the following:

51ujk9.png
 

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