Prove that x^n = e is a subgroup of G

Let G be an Abelian group with identity e and let n be some fixed integer.
Prove that the set of all elements of G that satisfy the equation
x^(n) = e is a subgroup of G. Give an example of a group G in which
the set of all elements of G that satisfy the equation x^(2) = e does not
form a subgroup of G.

G abelian means that members commute
д a,b є G| a.b = b.a є G where I am using the symbol д to mean ' for all '.

identity e є G | ea =a

defining x^n = x.x.x... n-times where assumed x є G above.

statement is :
x^n є H | H ≤ G (i.e. H is a sub-group of G) and x^n=e for a given n.

x itself is a subgroup of G iff x=e (unit member must exist) since e.e = e ...a trivial sub-group

now if x-exists, then x^n = x.x.x .. = e
x*x^n = (x*.x).x^(n-1) where x* is inverse element (whch also must exist since H is itself a group)
e.x^(n-1) = x^(n-1) and this also є H

so by repeated operation with x*.x^(n-1) we end up with e.x = x є H

or x, x^2, x^3 ... x^n constitutes a group H

but xє G also as given, so x.x= x^2 є G, x^3 є G...x^nє G
i.e. G is a group with members {H, y ≠ x} so H ≤ G


example:
let G| set Z and operation is addition (so x^2=x+x=2x) , 0=e

if x^2 = e => 2x =0 or x=e є G, but the only sub group is e-itself trivially
so no proper H exists!