Suppose that $\cdot$ is associative and has an identity element. Show that an element $g \in G$ has at most one inverse Let $(G,\cdot)$ be a group with $e$ its neutral element.
For an element $g\in G$, there exists one inverse element in $G$, denoted by $g^{−1}$, such that $g\cdot g^{−1}=g^{−1}\cdot g=e$. 
Can this be done any better?
And how can one show that $H:= H_1 \cap H_2$ is a subgroup of $G$, where $H_1$ and $H_2$ are subgroups of $G$?
Let $H=H_1\cap H_2$ where $H_1,H_2\leq(G,\cdot)$. Then, let $a,b\in H$
$\Rightarrow a,b\in H_1\wedge a,b\in H_2$
$\Rightarrow a\cdot b^{−1}\in H_1\wedge a\cdot b^{−1}\in H_2$
$\Rightarrow a\cdot b^{−1}\in H$
$H\leq G$
since $H\subseteq H_1$ and $H\subseteq H_2$.
Proof right?
Thanks
 A: a,b belongs to H,subset of G if ab^(-1) belongs to H then H is subgroup of G
H1 meet H2 is nonempty as 1 belongs to both of it
let a,b belongs to H1 meet H2 then ab^(-1) belongs to H1 and H2 { as H1 ,H2 are subgroups}
ab^(-1) belongs to H
hence h is subgroup
A: *

*Let $x', x''\in G$ be two inverses of $x\in G$.
We have the following:
$$x'=ex'=(x''x)x'=x''(xx')=x''e=x''.$$


The first equality is because $e$ is neutral element.
Second equality is because $x''$ is inverse of $x$.
Third equality is because of associativity.
The fourth equality is because $x'$ is inverse of $x$.
The fifth equality is because $e$ is neutral element.


*

*The following characterization of subgroups holds:



Let $(G,\cdot)$ be a group with neutral element $e$ and $H\subseteq G,\ H\neq\emptyset$. $H\leq G$ if and only if $\forall x,y \in H,\ xy^{-1}\in H$. 


The "only if" part is clear. For the "if" part:
Since $H\neq \emptyset \Rightarrow\exists a_0\in H$. 
From the hypothesis, we get that $a_0a_0^{-1}=e\in H$.
Furthermore, $\forall a\in H$, using the hypothesis, we get that $ea^{-1}=a^{-1}\in H$.
Applying again the hypothesis and the previous result, we get that $\forall a,b \in H: a(b^{-1})^{-1}=ab\in H$. Therefore, $\cdot|_{H\times H}:H\times H \rightarrow H$.
Associativity is inherited from $G$.
$H\neq\emptyset\Rightarrow\exists x\in H\Rightarrow x^{-1}\in H\Rightarrow xx^{-1}\in H\Rightarrow e\in H$.
Using this, we get that if $x,y\in H\Rightarrow x,y\in H_1\wedge x,y\in H_2\Rightarrow xy^{-1}\in H_1 \wedge xy_{-1}\in H_2\Rightarrow xy^{-1}\in H\Rightarrow H\leq G$.


*

*$G=\mathbb{R}\setminus\{-1\}$ 
$x \cdot y = x + y + xy$. 
Let us check that this is an operation on $G$, that means that if $x,y\in G$, then $x\cdot y\in G$. In our case, we would have to show that if $x,y\neq -1\Rightarrow x\cdot y\neq -1$. What does it mean that $x\cdot y =-1$?
$x+y+xy=-1\Rightarrow x+y+xy+1=0\Rightarrow (x+1)(y+1)=0$. Therefore if $x, y\neq -1$, $x\cdot y\neq -1$., which makes $\cdot $ an operation on $G$.


Now to check associativity, i.e. if $x,y,z\in G$, then $(x\cdot y)\cdot z=x\cdot (y\cdot z) $
$(x\cdot y)\cdot z= (x\cdot y) +z +(x\cdot y)z=x+y+xy+z+(x+y+xy)z=x+y+z+xy+xz+yz+xyz=x+(y+z+yz)+x(y+z+yz)=x+y\cdot z +x(y\cdot z)=x\cdot (y\cdot z)$.
$0$ is the neutral element. Indeed if $x\in G$: $x\cdot 0=x+0+x0=x=0\cdot x$.
To check for the inverse element: $x\cdot x^{-1}=0\Rightarrow x+x^{-1}+xx^{-1}=0\Rightarrow x^{-1}{1+x}=-x\Rightarrow x^{-1}=\frac{-x}{1+x}$.
I think you can compute your answer. Hope it is helpful.
