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[SOLVED] Herstein algebra hint(s) desired
Let G be an abelian group of order n (even), a_1 ,..., a_n its elements. Let x = a_1 \cdot ... \cdot a_n. Show that if G has more than one element b which is not the identity e, such that b^2 = e, then x = e.
(from Herstein, Abstract Algebra, section 2.4 problem 43b)
We know that x^2 = e.
We can assume that b = a_1 , a_2 and note that we must also have a third solution b = a_1 \cdot \ a_2 as well which differs from the first two. It would seem that a reasonable strategy would be to attempt to find a solution to cx = c, but I'm afraid I don't see it (and not for lack of experimentation).
Homework Statement
Let G be an abelian group of order n (even), a_1 ,..., a_n its elements. Let x = a_1 \cdot ... \cdot a_n. Show that if G has more than one element b which is not the identity e, such that b^2 = e, then x = e.
(from Herstein, Abstract Algebra, section 2.4 problem 43b)
Homework Equations
We know that x^2 = e.
The Attempt at a Solution
We can assume that b = a_1 , a_2 and note that we must also have a third solution b = a_1 \cdot \ a_2 as well which differs from the first two. It would seem that a reasonable strategy would be to attempt to find a solution to cx = c, but I'm afraid I don't see it (and not for lack of experimentation).

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