anyone please may help prove this abstract... thank you?

Question

Show that if G is a finite group with identity e and with an even numebr of elements, then there is a ≠ e in g such that a* a = e.

Answer 1

Look at G - {e}.
We know this has an odd number of elements.
Take any element a and pair it with its inverse.
If the inverse of a equals a, we are done.
Else multiply a by it's inverse and delete the pair.
There are still an odd number of elements left.
Now consider the remaining set and do the same thing.
Now G is finite so a finite number of steps we
will have (in the worst case) only one element left.
But now we can only pair a with itself, so
a = a^-1 or a*a = e.
In number theory, this is the same strategy
we use to prove Wilson's theorem:
If p is prime then (p-1)! = -1(mod p).
Let's look at U_11, for example
The elements are
1 2 3 4 5 6 7 8 9 10
Let's consider the elements 2, ...10 and do the above
pairing.
Working mod 11 we delete the pairs
(2,6), (3,4), (5,9) and (7,8)
Only 10 is left so 10 is its own inverse.
So if you multiply all the elements together
and use this strategy, their product is 10 = -1(mod 11).