This discussion focuses on proving the order of an element \( g \) in a finite group \( G \). The order \( o(g) \) is defined as the smallest natural number \( r \) such that \( g^r = e \), where \( e \) is the identity element of the group. Participants clarify that the set \( \langle g \rangle \) consists of distinct powers of \( g \) and must demonstrate that this set contains exactly \( r \) elements. The proof involves showing that any integer power of \( g \) can be expressed in terms of \( r \) and establishing that \( \langle g \rangle \) is equal to \( G_r \), which contains \( r \) distinct elements.
PREREQUISITESMathematicians, students of abstract algebra, and anyone interested in deepening their understanding of group theory and the properties of finite groups.
Bashyboy said:I am sorry to rehash this again, but I am having trouble with one argument I made:
I don't see why this implies that ##a## and ##k## have to have the same exponent. Isn't it possible that ##k > a##, yet ##g^k## and #g^a## get mapped to the same element?
Bashyboy said:I am not sure how I can conclude ##\langle g \rangle## and ##G_r## are the same set, then. Haven't I only demonstrated that ##G_r## is a subset?
Bashyboy said:Oh, and so this forces them to be equal, and it forces the exponents of ##\langle g \rangle##, without any repetition, to be {0,1,2,3...,r-1}. Is that right?