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I was looking at the proof of Lagrange's theorem (that the order of a group ##G## is a multiple of the order of any given subgroup ##H##) in Wikipedia:
I understand this proof fine, but I was wondering, instead of finding a bijection between cosets, is it enough to find a bijection between an arbitrary coset ##gH## and the subgroup ##H##? So, for instance, we have a map ##f: gH \rightarrow H##, where ##f(x) = g^{-1}x##. The map is bijective, with inverse ##f^{-1}(y) = gy##. Is there anything wrong with this?
This can be shown using the concept of left cosets of ##H## in ##G##. The left cosets are the equivalence classes of a certain equivalence relation on ##G## and therefore form a partition of ##G##. Specifically, ##x## and ##y## in ##G## are related if and only if there exists ##h## in ##H## such that ##x = yh##. If we can show that all cosets of ##H## have the same number of elements, then each coset of ##H## has precisely ##|H|## elements. We are then done since the order of ##H## times the number of cosets is equal to the number of elements in ##G##, thereby proving that the order of ##H## divides the order of ##G##. Now, if ##aH## and ##bH## are two left cosets of ##H##, we can define a map ##f : aH \rightarrow bH## by setting ##f(x) = ba^{-1}x##. This map is bijective because its inverse is given by ##f^{-1}(y)=ab^{-1}y##
I understand this proof fine, but I was wondering, instead of finding a bijection between cosets, is it enough to find a bijection between an arbitrary coset ##gH## and the subgroup ##H##? So, for instance, we have a map ##f: gH \rightarrow H##, where ##f(x) = g^{-1}x##. The map is bijective, with inverse ##f^{-1}(y) = gy##. Is there anything wrong with this?