Ways to deal with generating sets of groups I find that when I come across questions to do with generating sets of groups, I'm never quite sure how to go about the problem. It's difficult to deal with them purely set-theoretically, as you can't make definitive statements about what they contain (or can you?). I'm interested in how other people typically approach proofs involving generating sets, i.e. by contradiction, by definition, etc. Or perhaps it is better to approach them more abstractly, and attack them using theorems (isomorphism thms, correspondence thm, etc)? 
This question is obviously open-ended so to appease the monitors let us consider an example:

Let $S$ be a subset of a group $G$ and let $g\in G$. Show that $<S>$ satisfies $$g<S>g^{-1}=<gSg^{-1}>$$ 

We could perhaps use the definition, i.e. $$<S>=\bigcap \{H\leq G:S\subset H\}$$ But I didn't find that to help much. I also tried supposing that $\exists x\in g<S>g^{-1}$ such that $x\notin<gSg^{-1}>$ and vice-versa, but once again I found it to be quite fruitless. 
Comments and thoughts appreciated!
 A: In order to make "definitive statements" about what $\langle S \rangle$ contains, the key properties are:
(1) We have $S \subseteq \langle S \rangle$, and $1 \in \langle S \rangle$.
(2) If $x,y \in \langle S \rangle$, then $xy \in \langle S \rangle $ and $x^{-1} \in \langle S \rangle $.  
(3) If $H$ is any subgroup of $G$, and $S \subseteq H$, then $\langle S \rangle \subseteq H$.
Properties (1) and (2) are useful for proving that $\langle S \rangle$ contains something (a statement the form $\langle S \rangle \supseteq H$), while property (3) is useful for proving that $\langle S \rangle$ is contained in something (a statement of the form $\langle S \rangle \subseteq H$).
The following additional properties are useful for this problem:
(4) $\langle S \rangle$ is always a subgroup.  
(5) If $H$ is a subgroup of $G$, and $g \in G$, then $gHg^{-1}$ is a subgroup of $G$.
In your case, you want to prove that $$g \langle S \rangle g^{-1} = \langle gSg^{-1} \rangle$$ which is equivalent to the two statements $$g \langle S \rangle g^{-1} \subseteq \langle gSg^{-1} \rangle$$ $$g \langle S \rangle g^{-1} \supseteq \langle gSg^{-1} \rangle$$
So we want to prove that $\langle gSg^{-1} \rangle$ both contains and is contained in $g \langle S \rangle g^{-1}$.
To prove $g \langle S \rangle g^{-1} \subseteq \langle gSg^{-1} \rangle$: By (1), $$gSg^{-1} \subseteq \langle gSg^{-1} \rangle.$$    Multiplying $g^{-1}$ on the left and $g$ on the right, we have $$S \subseteq g^{-1} \langle gSg^{-1} \rangle g.$$  The right side is a subgroup by (4) and (5), so by property (3), $$\langle S \rangle \subseteq g^{-1} \langle gSg^{-1} \rangle g.$$  Multiplying $g$ on the left and $g^{-1}$ on the right, $$g \langle S \rangle g^{-1} \subseteq \langle gSg^{-1} \rangle.$$
To prove $g \langle S \rangle g^{-1} \supseteq \langle gSg^{-1} \rangle$: By property (1), $$S \subseteq \langle S \rangle.$$  Multiplying $g$ on the left and $g^{-1}$ on the right, we have $$gSg^{-1} \subseteq g\langle S \rangle g^{-1}.$$  The right side is a subgroup by properties (4) and (5), so by property (3), $$\langle gSg^{-1} \rangle \subseteq g\langle S \rangle g^{-1}.$$
A: The group generated by a subset can be constructed, as a set, by forming all finite compositions of (powers of) elements of the subset and their inverses. In your example that is helpful information because the intermediate $g$'s cancel out.
