show that every group G with identity e such that x^2=e for all x is an element of G?

Question

I WANT TO KNOW MORE ABOUT GROUPS IN ABSTRACT ALGEBRA

a*a=e
b*b=e
THIS IS FROM ABSTRACT ALGEBRA

Answer 1

"A group is a set of objects with a binary operation (usually called a multiplication)". To form a group, the set has to be "closed" under the operation: use the operation on any two members of the set (including the same member used with itself) and the result is a member of the set). It also needs an "identity", that is a member such that if the operation is between any member and the identity, the original member is left unchanged. Also, each member must have an inverse (another member such that the multiplication of a member by its inverse gives the identity. The associative law must hold
(a * b) * c = a * (b*c) meaning the order doesn't matter.

e.g., the set of all even numbers forms a group under addition. take any two even numbers, add them together and you get an even number. 0 is an even number which, when added to another even number, will leave it unchanged. Therefore, 0 is the identity in the group (Even, +). For every even number x, there exists an even number -x such that adding x and -x gives the identity (0).

Whatever the group operation, it is normally called a multiplication. The symbol for the identity can be 1, I, e or any other symbol you need to use to ensure that others will know this is the identity (in groups where members include the complex i -- root of -1 -- you cannot use the letter i for the identity). Here, I think that you are using the letter e to identify the identity (not the value 2.718281828459...).

The word addition is only used when it is clear that you mean the arithmetical addition; often you do this if you then plan to turn the group into a ring (where you have two operations and two identities: a zero for addition and a one for multiplication). When the group uses the addition, the symbol for the identity is often 0.

As for your question, I'm not sure what you mean by "every group G ... is an element of G"

What you describe in the first part, is a group where any element is a root of the identity.

For example, take a set that contains only the number 1 and apply the arithmetical multiplication, then you have that every element is such that x^2 = 1.

Klein devised a group of four elements; a, b, c and 1, such that:

a^2 = 1
b^2 = 1 but b is not equal to a (therefore b*a is not equal to 1)
c^2 = 1 but c is not equal to a nor to b.
and
a*b*c = 1

From that, it is possible to prove that the set is closed under the operation, that 1 is the identity and that every member has an inverse (itself). In this case, the group is also Abelian (ab = ba).

Answer 2

What you want to show is actually that in a group G with identity e, if for all x in G x*x=e, then G is commutative, i.e. x*y=y*x for all x, y.
But if (x*y)*(x*y)=e, that means that x*y*x*y=x*x*y*y, so y*x=x*y.