- #1
pedrommp
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Homework Statement
If [tex] G [/tex] is a group in which [tex] (a.b)^i=a^i.b^i [/tex] for three consecutive integers [tex] i [/tex] for all [tex] a,b \in G [/tex] , show that [tex] G [/tex] is abelian.
Show that the conclusion does not follow if we assume the relation [tex] (a.b)^i=a^i.b^i [/tex] for just two consecutive integers.
Homework Equations
The basic group identities
The Attempt at a Solution
Well, the first part i made :
Let j be the smallest of the three consecutive integers. Then we have that
[tex] (a.b)^j = a^j.b^j [/tex]
[tex] (a.b)^{j+1} = a^{j+1}.b^{j+1} [/tex] and
[tex] (a.b)^{j+2} = a^{j+2}.b^{j+2} [/tex] for all [tex] a, b \in G [/tex]
Inverting the first equation and right multiplying it to the second equation implies
[tex] a.b = a^{j+1}.b^{j+1}.b^{-j}.a^{-j} = a^{j+1}.b.a^{-j} [/tex] hence
[tex] b.a^j = a^j.b [/tex]
Inverting the second equation and right multiplying it to the third equation gives
[tex] a.b = a^{j+2}.b^{j+2}.b^{-j-1}.a^{-j-1} = a^{j+2}.b.a^{-j-1} [/tex] hence
[tex] b.a^{j+1} = a^{j+1}.b [/tex]
Therefore [tex] a^{j+1}.b = b.a^{j+1} = b.a^{j}.a = a^{j}.b.a [/tex] and left multiplying by
[tex] a^{-j} [/tex] yields [tex] a.b = b.a [/tex] for arbitrary [tex] a, b \in G [/tex]
Hence G is abelian.
But the second i tried to find an example of non-abelian group that satisfies the relation for two consecutive integers with no success.