I've been studying about subgroups and I encountered an example with answers and does not have explanation how it is derived and I need help to understand it.
Here is the example:
Example 1.4.20
Determine whether the given subset of the complex numbers is a subgroup of the group $\mathbb{C}$ of complex numbers under addition.
a.) $\mathbb{R}$: YES
b.) $\mathbb{Q}^+$: NO, there is no identity element.
c.) $7\mathbb{Z}$: YES
d.) The set of $i\mathbb{R}$ of pure imaginary numbers including $0$: YES
e.) The set $\pi\mathbb{Q}$ of rational multiples of $\pi$: YES
Best Answer
For a subset to be a subgroup, it has to be closed under the group's binary operation and the formation of inverse. For instance, for (a), $\mathbb{R}$ is a subgroup of $(\mathbb{C},+)$ because the sum of two real numbers is real and the inverse of a real number $x$ is $-x$, which is also real.
For (b), the answer is no because the identity element of the group is $0$, and it does not belong to $\mathbb{Q}^+$. Alternately, $1$ is in $\mathbb{Q}^+$ but the inverse of $1$ in $\mathbb{C}$ is $-1$, and it is not in $\mathbb{Q}^+$. So $\mathbb{Q}^+$ is not closed under taking of inverse, and is not a subgroup.