# Finite Cyclic Group

• Jan 26th 2013, 03:44 PM
lovesmath
Finite Cyclic Group
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!
• Jan 26th 2013, 10:27 PM
Deveno
Re: Finite Cyclic Group
a word about an element of order 2: if |x| = 2, then x2 = e. that is, xx = e, so x must be its own inverse.

on the other hand if x = x-1, then x2 = 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 xk. it is clear that |xk| = 2, since xk ≠ e (k < 2k, and 2k is the SMALLEST positive power of x that equals e), and (xk)2 = x2k = e.

for suppose we had y ≠ e with y2 = e. since G is cyclic, y = xm, for some 1 ≤ m ≤ 2k-1.

since |y| = 2, y2 = (xm)2 = x2m = 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 = y2 = (xm)2 = (xk+s)2 = x2k+2s = (x2k)(x2s) = e(x2s) = x2s.

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,x2,x4,....,x2k-2 can be e), that is: s = 0, so in fact y must be xk.
• Jan 31st 2013, 07:59 AM
lovesmath
Re: Finite Cyclic Group
This was extremely helpful. Thanks!