$x=x^2$ in a sub group? I have a set E defined in ℝXℝ (E=ℝXℝ) and the operation *
defined like this 
(a,b)*(c,d)=(a+c,b+d+2ac)

(a,b),(c,d) are elements of E, I want to   


*

*prove that (E,*) is a Group.

*prove that the set H defined like so H={(x,x), x ∈ ℝ} is a Sub Group of (E,*).

*we define the function f like so : 
   f: (ℝ, + )→ (H, * )                    
       x     → f(x)=(x,x²)

prove that f is a Group homomorphism.
-To prove that E is Group, I must prove its Closure that's mean 
For all (a, b), (c,d) in E, the result of the operation, (a,b) * (c,d), is also in E (verified).
Associativity
For all (a,b), (c,d) and (e,f) in E
 ((a,b) * (c,d)) * (e,f) = (a,b) * ((c,d) * (e,f)).

and that's mean :
((a,b)*(c,d))*(e,f)=(a+c,b+d+2ac)*(e,f)=(a+c+e,b+d+2ac + f + 2(a+c)e) 
==
(a,b)*((c,d)*(e,f))=(a,b)*(c+e,d+f+2ce)=(a+c+e,b + d+f+2ce + 2(c+e)a)

(verified)
Identity element
There exists an element (e1,e2) in E, such that for every element (a,b) in E, the equation (e1,e2) * (a,b) = (a,b) * (e1,e2) = (a,b) holds.
if we develop we get  :
{a+e1==a     => e1=0
{b+e2+2ae1=b    =>e2=0 (using e1=0)

so (e1,e2)=(0,0)
Inverse element:
let (a1,b1) be the Inverse element of (a,b) 
(a1,b1)*(a,b)=(a,b)*(a1,b1)=(0,0)
{ a1+a=0 =>a1=-a
{ b1+b+2a1a=0 =>b1=2a²-b

so the inverse element of (a,b) is (-a,2a²-b)
The Problem 
now the problem occurred when i tried to prove that (H, * ) is a Sub Group of (E, * ) because to do that you must prove that whatever element (a,a) in H, its Inverse element (a',a') must also be in H, but as you can see that the Inverse element of (a,a) is (-a, 2a²-a) which is clearly not an element of H    ?
And if you take a look at the Question 3 you can see that f(x)=(x,x²) is supposed to be an element of H (f goes from E → H )! and that's mean that x²=x !!
I exclude the possibility that the question it self is Wrong, am i missing something here ?
..any help is appreciated. 
 A: The set $H=\{(x,x)\}$ isn't closed under the group operation:$(s,s)*(t,t)=(s+t,2st+s+t).$
However if we reset $H$ to be the pairs $(x,x^2)$ it is closed, i.e. with this reset the group operation gives $(s,s^2)*(t,t^2)=(s+t,(s+t)^2).$ So if you work with this other definition of $H$ you can likely find the inverses also.
Note that from the third part of the question, about a homomorphism from $R$ to $H$ given by $f(x)=(x,x^2)$ it would seem the question setup was for the $H$ to consist of pairs $(x,x^2)$.
A: This just addresses the question of showing that your formulas give you a group. It may well be that you don’t yet have the background for getting this on your own, but it does represent a much quicker route to the desired result.
One would be awfully surprised if this proposed group structure could not be represented as matrices. Indeed, after just a little experimentation, I saw that you could represent $(a,b)$ as the matrix
$$
\pmatrix{1&a&b\\0&1&2a\\0&0&1}\,.
$$
You multiply two,
$$
\pmatrix{1&a&b\\0&1&2a\\0&0&1}\pmatrix{1&c&d\\0&1&2c\\0&0&1}
=
\pmatrix{1&a+c&b+d+2ac\\0&1&2(a+c)\\0&0&1}\;
$$
just what you wanted it to be. Finally, you calculate the inverse of the matrix with $a$ and $b$ in the first row, and find that it is of the right form, and its first row entries are $-a$ and $-b+2a^2$. You have a group because your special matrices form a subgroup of the group of matrices of determinant $1$, done.
