Is Rational number under multiplication group? A binary operation $*$ is defined on $\mathbb{Q}$ such that $a*b=ab$.
So, it is just simple multiplication. 
I found out that this is a group with identity $=1$ and inverse is reciprocal of element "$a$" in $G$.
But, my answer key says that this is not a group because axiom 3 (inverse) fails.
I can't see why it is not a group.
Is my answer key wrong or am I wrong? 

HINT: What is the multiplicative inverse of $0$?
If we replace $\mathbb Q$ with $\mathbb Q\setminus \{0\}$ to exclude $0$ from our set, under the operation of multiplication, we do then have a group: $(\mathbb Q\setminus \{0\}, \cdot)$
