MHB Basis Theorem for Finite Abelian Groups

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The discussion focuses on proving that if the equation a^{l_0}b_1^{l_1}...b_n^{l_n}=e holds, then each term must equal the identity element e. The user has successfully completed earlier parts of the problem but struggles with this final proof. They suggest that if the order of a is strictly greater than that of each b_i, they could use induction to show a^{l_0}=e. However, they express uncertainty about proving the strict inequality of the orders of a and b_i. The conversation emphasizes the need for a deeper understanding of group orders in 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:

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