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?).
Best Answer
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 $(1)(1) = (-1)(-1) = 1$.