abstract algebra?

Question

prove that agroup of order 12 must have an element of order 2

Answer 1

Let G is a group with neutral element e, |G| = 12. The order of every element of G is a divisor of 12, so it can be in principle 1, 2, 3, 4, 6, or 12, the only element of order 1 is e.
If x is element of order 12 that means all its degrees
x, x², x³, . . , x^11, x^12 = e
are distinct /G is a cyclic and x is a generator/, then
x^6 ≠ e and (x^6)² = e, so |x^6| = 2.
If G does not have elements of order 12, but, say y is element of order 6, the same approach leads to (y³)² = e, so |y³| = 2.
If there is no such y, but z is of order 4, similarly |z²| = 2.
Assuming no elements of orders 2, 4, 6 and 12 in G it remains G contains e and 11 elements of order 3. But the latter is impossible: if |u| = 3 then u, u², u³ = e are distinct, so |u²| = 2 and the number of elements of order 3 would be even - contradiction.
The above is true for every group of even order - the number of elements of odd order (3 or more) is even, and if x is an element of even order either |x| = 2, or if x^(2n) = e /n=2,3,../ then |x^n| = 2.

Answer 2

It's possible to give a simpler proof of the general result:

Any group of even order has an element of order 2.

Why?

Lemma: Let G be of finite order. The number of elements of G order greater than 2 is even.

Proof: Pick such an element. It doesn't equal it's inverse. Pair them up. Keep going until you've exhausted all the elements of order >2. Q.E.D.

The order of G is 1 (for the identity) + the number of elements of order 2 + the number of elements of G of order >2.

So if |G| is even, the number of elements of order 2 is odd, so in particular it doesn't equal 0.

Q.E.D.!