What about , i.e., 5-tupes of 0, 1 under addition?
People in the Algebra forum may be better equipped to answer this question. I would ask a moderator to move this thread.
True of false - for any group G, the number of elements x belongs to G such that x^4 = e is at most 4.
I found on another forum the following solution but many steps were skipped...
"Clearly e^4 = e.
x^5 is an element such that (x^5)^4 = e.
So are x^10 and x^15.
But these are just 4 elements.
So clearly there is no counterexample with a cyclic group.
Now you try examples with non-cyclic Abelian groups, and Non-Abelian groups and see what you come up with..."
what actually is C_20??and i cant think of an example of any non-cyclic abelian groups can anyone please talk me through where is method is going, preferably step by step??thanksss
here is a non-abelian group of order 8, the 4th dihedral group D4 = {1,r,r^2,r^3,s,rs,r^2s,r^3s} with r^4 = s^2 = 1, sr = r^3s.
we have the subgroup of order 4, {1,r,r^2,r^3} and clearly, for every element x in that subgroup, x^4 = 1.
(1^4 = 1, r^4 = 1, (r^2)^4 = (r^4)^2 = 1^2 = 1, (r^3)^4 = (r^4)^3 = 1^3 = 1).
so there are 4 elements right there. but s^4 = (s^2)^2 = 1^2 = 1, so s is a 5th element whose 4th power is 1.
when looking for counter-examples in groups, non-abelian groups of small order often work well.