$G = \left\{1, -1\right\}$ is a group under multiplication I'm reading about groups for the first time, and I'm a bit confused as to how to prove that $G = \left\{1, -1\right\}$ is a group under multiplication.
(A) $1, -1 \in G $ implies $(1)\cdot(-1) = -1 \in G $ (closure). 
(B) For any $1, -1 \in G $ we have $-1 = (1)\cdot(-1) = (-1)\cdot(1) = -1$ (associativity). 
(C) There exists $1 \in G$ such that for $-1 \in G$ we get  $(1)\cdot (-1) = (-1)\cdot (1) = -1 \in G$ (identity element). 
(D) For any $-1 \in G$ exists $-1 \in G$ such that $(-1)\cdot (-1) = (-1)\cdot(-1) = 1 \in G$ (inverse element). 
However, I'm pretty sure this is incorrect (the last one doesn't make sense cause I'm using the same element twice?). 
 A: It's mostly fine, but in all cases, you have to check for every element, not just ones that are distinct. For example, to check it is closed under multiplication, you should also verify that $(1)(1) = (-1)(-1) = 1$. 
A: If you know $\{ 0, 1 \}$ is a group under addition $\mod 2$, then you can show yours is by the isomorphism $\phi(x) = (-1)^x$.
A: We need to fix a few things.
(A) Closure: you need show that for all elements $a,b\in G$ we have $ab\in G$. So, we need to check that $1\cdot 1, 1\cdot (-1), (-1)\cdot 1, (-1)\cdot (-1)\in G$.
(C) Identity: There exists $1\in G$ such that for all $a\in G$ we have $1\cdot a=a\cdot 1=a$. So, we need to show that $1\cdot 1=1\cdot 1=1$ and that $1\cdot(-1)=(-1)\cdot1=-1$.
A: For associativity, you should by definition show $1\cdot (1\cdot (-1))=(1\cdot 1)\cdot (-1)$ and so on, all the 8 possibilities. Their number can be reduced by various considerations. For example, you can argue that the multiplication coincides with multiplication of real numbers and $a\cdot (b\cdot c)=(a\cdot b)\cdot c$ holds for real numbers.
For the other axioms, as other have pointed out, you also need to check them for all elements, not only for one. For example, for (D), not only $-1$ is inverse to $-1$ but also $1$ is an inverse to $1$.
