Introduction to groups - question/proof check Prove if the following are abelian groups, if so, list the neutral element e:
(a) $G:= \Bbb Z, a*b:= max \{a,b\}$ (Z as in positive integers incl. zero (i dont know how to write it with mathjax)) 
(b) $G:= \Bbb R, a*b:= a^b$ (only positive real numbers)
(c) $G:= \Bbb Z, a*b:= lcm(a,b)$ (least common multiple of a and b) (integers greater than zero)
(d) G is the open interval $(0,1) \subset \Bbb R $ with
$$x*y: = \frac{xy}{1-(x+y)+2y} $$
Hi guys, the university only gave me some definitions, no further explaination whatsoever, im not even sure i understand this correctly (uni is really terrible at teaching)
First, take example a), I assume "$G:= \Bbb Z, a*b:= max \{a,b\}$ " means a group G with the elements a and b $\subset \Bbb Z $. So i need to use the 4 definitions to prove this to be a abelian group.
EDIT:
2) (b) $G:= \Bbb R, a*b:= a^b$ 
Associativity: $\forall a,b,c \in G: a\circ(b\circ c) = (a\circ b)\circ c$ For question b would mean $a^{b^c}=a^{b^c}$ which is correct...
Invertibility:$\forall a \in G \exists a' \in G: a' \circ a = e$ Suppose its invertible -> $a'\circ a=e \rightarrow a^{-1^{a}}=e$ However, i do see from my schools paper that $a \circ a^1=e$ should also hold, but then in this case that would be $a^{a^{-1}}$ which is not the same thing, so invertibility also fails?
Identity: $\forall a \in G: e \circ a = a$ Since invetibility fails i cant say much about identity...
Commutativity: $a \circ b = b \circ a$ $\rightarrow a^b=b^a$, unless a=b=1, that is not true either...
If someone has time to look it over and give me some suggestions or point out where is wrong or how to write the proofs more formally id be really thankfull!!!!
 A: I think you are confusing the operation (which is often called multiplication for Groups). When we say we multiply 2 elements $a$,$b$ in a group, we mean we do the group operation "$\circ$" for $a$ and $b$. That is, $a \circ b$.
For example, The real numbers form an abelian group $G$, with operation $a \circ b = (a+2)+(b+2)$ (regular addition as a part of the operation)

*

*Associativity: $(a \circ b) \circ c = \left((a+2) + (b+2)\right) \circ c = (a+b+4) \circ c = (a+b+4)+(c+2) = a+b+c+6 = a \circ (b+c+4) = a \circ \left((b+2)+(c+2) \right) = a \circ (b \circ c)$.

*Identity: Does there exist an element $e$, such that $a \circ e = a$? Let $a$ be chosen randomly from the elements of $G$, and let $x=-4$. Notice that $a \circ x = (a+2) + (x+2) = (a+2) + (-4+2) = a$. Thus we can deduce that $e=-4$.

*Invertibility Does there exist an element $a^{-1}$, such that $a \circ a^{-1} = e$? This is the same as solving $(a+2)+(a^{-1}+2)=-4$, which gives us $a^{-1}=-a-8$. Once more, since we can choose $a$ randomly (or "arbitrarily") from the elements of $G$, that means all elements of $G$ has an inverse!

It can also be shown that our group $\left(G,\circ\right)$ is commutative, however I don't want to put it there, because it is not necessary to form a group. The one thing I want to emphasize is that we are not multiplying (unless the operation is a literal multiplication) anything, but rather using the group operation. In your question, you apply the group operation first, then multiply it again.
If you would like, I can also point out why/why not (a),(b),(c),(d) are groups.
