Let G be a finite cyclic group. Show that G has exactly one element of order two if and only if |G| is even. Please help!
a word about an element of order 2: if |x| = 2, then x^{2} = e. that is, xx = e, so x must be its own inverse.
on the other hand if x = x^{-1}, then x^{2} = x(x^{-1}) = e. this means |x| = 1 or 2, so if x is not the identity, it means |x| = 2 (only the identity has order 1).
so for a non-identity element of G, having order 2 and being its own inverse are the same thing.
now for ANY finite group G, suppose we have just ONE element g of order 2. then (aside from this element, g, and the identity, e), we can list all the elements of G in pairs:
x,x^{-1}
y,y^{-1}
.....
etc.
so |G - {e,g}| must be even, say |G - {e,g}| = 2k. hence |G| = 2k+2, which is even. since this is true for ANY finite group, it is certainly true for CYCLIC finite groups.
for the reverse implication, we must assume G is cyclic (there exist non-cyclic groups of even order with more than one element of order 2). so suppose |G| = 2k, and G is cyclic.
let x be a generator for G (there is one, since G is cyclic). thus |x| = 2k.
i claim the only element of order 2 in G is x^{k}. it is clear that |x^{k}| = 2, since x^{k} ≠ e (k < 2k, and 2k is the SMALLEST positive power of x that equals e), and (x^{k})^{2} = x^{2k} = e.
for suppose we had y ≠ e with y^{2} = e. since G is cyclic, y = x^{m}, for some 1 ≤ m ≤ 2k-1.
since |y| = 2, y^{2} = (x^{m})^{2} = x^{2m} = e.
since the order of x is 2k, we must have 2k divides 2m. thus k divides m, so k ≤ m. so let's write m = k + s. clearly 0 ≤ s ≤ k-1.
so e = y^{2} = (x^{m})^{2} = (x^{k+s})^{2} = x^{2k+2s} = (x^{2k})(x^{2s}) = e(x^{2s}) = x^{2s}.
since 0 ≤ 2s ≤ 2(k-1) ≤ 2k-1, we have to have 2s = 0 (because the order of x is 2k, so none of the powers x,x^{2},x^{4},....,x^{2k-2} can be e), that is: s = 0, so in fact y must be x^{k}.