Thread: (LaGrange) Proving/disproving that G; |G|= x^y is either cyclic or a^m=e for all a∈G

The following likely requires LaGrange's Theorem / cosets, but I am unsure how to construct the necessary counterexamples, or prove/disprove part (b). Any help/hints/tips on any of this would be very much appreciated.

We know that if G is a group of order 4 with identity e, then either G is cyclic or a² = e for all a∈G.

(a) Find a counterexample to prove that this result cannot be generalized to use a base that is a natural number m other than 2. That is, disprove the following proposition:
If m is a natural number greater than 2 and G is a group of order m² with identity e, then G is either cyclic or a^m = e for ever a∈G.


(b) Is there any condition on m that will make the proposition in (a) true? If yes, state and prove a conjecture. If no, explain why.


(c) Does the stated result generalize to other powers of 2? That is, prove or disprove the following proposition.
If G is a group of order 2^k for some k∈ℤ with k>2, then either G is cyclic or a² = e for every a∈G.

G is a group of order m and m> 2, with m prime number
let and x differ from the identity
from lagrange theorem, the order of x divide the order of G
so x order is m^2 or m
if it is m^2 then G is generated from x which means G is cyclic
then other case x order is m.

Thanks Amer, that helps tremendously with part (b).

I am still working to construct counter examples for (a) and (c), any advice on this?

the following counter-example works for both a) and c):

let G = Z8 x Z2, which is of order 16 (= 4^2 = 2^4).

consider the element (1,1) in G. this element is of order 8.

Thanks for that, however, I should have mentioned that we have only worked with Z_x, Dihedral, and U groups. Are there any counterexamples within these groups?

look at U(32)