Abstract Algebra : Subgroups 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
 A: 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.
A: Here's a (maybe too verbose) explanation:
A subgroup is a special subset of a group, specifically it's special because it forms a group in its own right (under the same operation as the group containing it). 
Example: We know, or can quickly check that $\mathbb{C}$ (the complex numbers) is a group under addition. I'm not sure what you're Prof.'s favorite version of the group axioms are, but here's one version of them:


*

*The operation $+$ is associative: $a+(b+c)=(a+b)+c$ i.e., it doesn't matter if we add $b$ and $c$ first or $a$ and $b$ first. 

*$\mathbb{C}$ is closed under addition, adding any two complex numbers is going to give you another complex number

*The number $0$ acts as the identity element ($0+x=x$ for any $x\in\mathbb{C}$).

*Every number $x$ has an additive inverse, namely  $-x$.


(Note that this list isn't as short as it could be, but I think it's about right for someone just learning groups)
We know that the reals are contained in $\mathbb{C}$, so $\mathbb{R}$ is a subset of $\mathbb{C}$, but it's a subset which also satisfies these four axioms of its own. 


*

*Addition is associative comes free from the fact that it's associative in $\mathbb{C}\supset\mathbb{R}$

*$0$ is a real number, so the identity is inside $\mathbb{R}$. 

*If you add two reals, you get a real so closure under the operation holds, and 

*If you negate an element of $\mathbb{R}$, you get another element of $\mathbb{R}$, so every inverse of an element of $\mathbb{R}$ is also in $\mathbb{R}$. 


By contrast, the group of positive rational numbers $\mathbb{Q}^+$ is not a subgroup of $\mathbb{C}$ because it does not contain the identity element. ($0$ isn't positive) 
Hopefully that's a good start, if there are other examples on the list that might help, please say so in comments.
Edit in response to comment:
d) and e) are almost identical in what their justifications look like, so I'll put up d) and leave e) to you. 
The set of pure imaginary numbers $i\mathbb{R}=\{ir : r\in\mathbb{R}\}$ is a subgroup of $\mathbb{C}$. 


*

*Associativity (as always) comes as a freebie from the associativity of $+$ on $\mathbb{C}$. (Honestly, you can probably after a while drop this from subgroup questions because it never fails)

*The pure imaginary numbers are closed under addition. Suppose we have $ir_1$, $ir_2\in i\mathbb{R}$. Then $ir_1+ir_2=i(r_1+r_2)$. Since $r_1+r_2\in\mathbb{R}$, $i(r_1+r_2)\in i\mathbb{R}$.

*The inverse of a pure imaginary number is also a pure imaginary number. Take $ir$ for some $r\in\mathbb{R}$. Then $-r\in\mathbb{R}$ and $-ir\in i\mathbb{R}$. 

*We're given that the identity is in $i\mathbb{R}$, but it probably wouldn't hurt for you to point out why.  

