The meaning of 'integral powers' and how $\{2,3\}$ generates $\mathbb Z_6$ There is a line in my text 

This shows that all such products of integral powers of $a$ and $b$ form a subgroup of $G$...

What does 'integral powers' mean here?
Additionally, there is an example that says $\{2,3\}$ generates $\mathbb Z_6$ since 

$2+3 = 5$, so that any subgroup containing $2$ and $3$ must contain $5$

I understand that if the set contains $5$ it has to generate $\mathbb Z_6$, but why do you take the step $2+3$?
The notation $\{2,3\}$ suggests that you take powers of $2$ under $+_6$ and separately take powers of $3$ under $+_6$.
In other words, is it ok to do the following?
$\{2,3,4\}$ generates $\mathbb Z_6$ since


*

*$2+_63 = 5$

*$3+_64 = 1$

*$2+_64 = 0$


and hence the set $\{2,3,4\}$ generates $\mathbb Z_6$? 
Also, is it incorrect to say that 'the product of the integral powers of the elements of the set $\{2,3,4\}$ generates $\mathbb Z_6$?
 A: I suppose you are considering the additive group $\mathbb{Z}_6$. The notation 
$\{2, 3\}$ is just the standard notation for a set. Now, a subset $S$ of a group $G$ generates $G$ if every element of $G$ is a product of elements of the form $s$ or $s^{-1}$, where $s \in S$. If the group is commutative, and if $S = \{a, b\}$, as in your example, it is equivalent to saying that every element of $G$ can be written as $a^{n}b^m$ for some $n, m \in \mathbb{Z}$ (I suppose that the sentence products of integral powers of $a$ and $b$ refers to this fact).
In additive notation, this would translate to $na + mb$. For instance, the subgroup of $\mathbb{Z}$ generated by $\{2, 3\}$ is formally the set of integers of the form $2n + 3m$, where $n, m \in \mathbb{Z}$. In this case, it is not difficult to see that this group is $\mathbb{Z}$ itself. Similarly, consider the additive group $(\mathbb{Z}_6, +_6)$ and the subgroup $H$ generated by $\{2, 3\}$. According to the general definition, $H$ is the set of elements of the form $2n +_6 3m$, where $2n$ stands for 
$$
\underbrace{2 +_6 \dotsm +_6 2}_\text{n times}.
$$
In particular, $H$ contains $2 +_6 3 = 5$ (take $n = m = 1$). And since $5$ generates $\mathbb{Z}_6$, $H = \mathbb{Z}_6$.
A: You must take 2+3 because it is describing a subgroup, and requires closure under operation as per the group axioms. If 5 wasn't there, it wouldn't be closed.
The notation implies taking all possible powers of both, and counting any duplicates once. So you get 2, 4, 0 and 3, 0 - but only count 0 once, so the whole set is {0, 2, 3, 4}
You don't need 4 in the generating set, because while 2+2 generates 4 and 4+4 generates 2, we only need one, and it's easier to throw away the 'largest' or 'higher' element.
