about abstract algebra topics on groups?

Question

1. show that every G with an identity e and such that x*x = x for all x element G is abelian. [hint: consider (a*b) * (a*b)]

2. show that if G is a finite group with identity e and with an even number of elements, the there is a not equal to e in G such that a*a=e

please how your complete answer or proofs. thanks!

Answer 1

1. Actually, not only is the group abelian, but
it consists only of the identity element.
To see this, note that
x*x = x.
Now multiply both sides on the left by x ^-1 to get
(x^-1*x)*x = x^-1 * x = e,
which gives
x = e.


2. Look at S = G- {e}.
Now pair every element of S with
its inverse. Since S has an odd number
of elements and every pairing uses up
2 elements for those elements which are
not self inverses, there must be at least
1 element a in S left over which is its own inverse.
In number theory, that's the strategy used
to prove Wilson's theorem.
For example, let's look at Z_11 under multiplication.
The elements are
1,2,3,4,5,6,7,8,9 and 10.
The pairings are
2 6
3 4
5 9
7 8
which leaves only 10 as a self inverse element.