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m0bius
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Homework Statement
Let G be a group and let a be an element in G. Show that a and a-1 generate the same cyclic subgroup <a> = <a-1> and have the same order o(a) = o(a-1).
Homework Equations
This is in the second section of the book I'm using, so there aren't really any theorems that can be used, just the definition of a group, order, and cyclic groups.
The Attempt at a Solution
First we show o(a) = o(a-1).
Assume a =/= e, and that both o(a) and o(a-1) are finite.
Let o(a) = n and o(a-1) = m. Then an = e and (a-1)m = e by the definition of order. Since a =/= e, m > 1 and n > 1. Multiply these two equations together to get:
(an)((a-1)m) = e. (1)
Case 1 : Assume m > n. Then we have (a-1)m-n = e. This can be seen by expanding (1), then canceling the a's with the a-1's from the inside outwards. Since n > 1, we can say m > 2. But then 0 < m - n < m, and so m is not the least positive integer such that (a-1)m = e. So o(a-1) =/= m and this is a contradiction.
Case 2 : Assume m < n. Then we have an-m = e, which can be seen by expanding and canceling like we did in Case 1. Since m > 1, we can say n > 2. But then 0 < n - m < n, and so n is not the least positive integer such that an = e. So o(a) =/= n and this is a contradiction.
Since m is not greater than n, and n is not greater than m, we must have n = m, and so o(a) = o(a-1).
I think I made it longer than it needed to be, but it seems correct. Anyway, my problem is now with the case when either or both of o(a) and o(a-1) are infinite. Let me just say I have about 5 pages or so of frantic scribblings trying to figure out what would happen in this case, and I am now extremely confused. My book only mentions one example at this point of an element with infinite order : all n =/= 0 in {Z, +} have infinite order (I understand this).
I'm not sure if it even makes sense to say that two elements of infinite order have the same order, but I'm currently attempting to show that if o(a) is infinite then o(a-1) is infinite. Any help/hints would be appreciated.
Thanks,
m0bius.