If $S$ is an Infinite Set, and $f(S)\subseteq{G}$ is a Set that Generates a Group $G$, then the Cardinality of $G$ is $\le$ Cardinality of $S$? As the header says, if we map an infinite set $S$ into $G$ by $f$, such that $f(S)$ generates $G$, then is $|G|\le|S|$? I know that this should be true in the case where $S$ does not include into $G$. Then it is obvious. What I'd like to show is that if $S$ does include into $G$ under $f$, then it remains true. 
I believe that is fairly straight forward in the case that $|f(S)|=|\mathbb{N}|$, since then we can enumerate the elements of $f(S)$ in such a manner that we can construct a bijection from $f(S)$ to $\mathbb{N}$, say $F\colon f(S)\to\mathbb{N}$. Then it is clear by induction that there is a bijection from $\mathbb{N}$ to $\mathbb{N}^n$ for all $n\in\mathbb{N}$. Thus, if we describe the order in which we operate on things in $F(S)$ by these ordered tuples, then we are looking at $\mathbb{N}^n$ for some $n\in\mathbb{N}$, and of course, all the $n$ are finite, and if $G$ is generated by $f(S)$, then we are looking at finite operations that are ordered. How would I show this for $|S|$ any size of infinity? It feels intuitively obvious that for any infinity $|S|^{n}=|S|$ for all $n$. I should note that I've never done transfinite induction, but I know what it is, in the sense that it is induction and it involves ordinals...

The easiest way to go about this would probably be to construct a surjection from a set with cardinality $|S|$ to $G$. To that end, let
$$X = \bigsqcup_{n \ge 0} (S \times \{\pm 1\})^n$$
be the disjoint union of powers of $S \times \{\pm 1\}$. This has cardinality $$|X| = \sum_{n \ge 0} |(S \times \{\pm 1\})^n| = \sum_{n \ge 0} |S| = |\mathbb{N}| \cdot |S| = |S|$$ (because $S$ is infinite, $|S| = |S \times \{\pm 1\}| = |(S \times \{\pm 1\})^n|$). The surjection $g : X \to G$ is given by
$$g((x_1, \epsilon_1), \dots, (x_n, \epsilon_n)) = f(x_1)^{\epsilon_1} \dots f(x_n)^{\epsilon_n}.$$
The fact that $g$ is a surjection is precisely a translation of the fact that $S$ is a generating set.
