Generator of a group Let $G$ and $H$ be two groups and let $f:G\longrightarrow H$ be an isomorphism.
Are the two following properties true?
1) If $g$ is a generator of $G$ then $f(g)$ is a generator of $H$ and more generally if $\langle g_1,\dotsc,g_n | R_G \rangle $ is a presentation of $G$, $R_G$ being a set of relations, then $\langle f(g_1),\dotsc,f(g_n) | f(R_G) \rangle $ is a presentation of $H$.
2) If $h$ is a generator of $H$, then $f^{-1}(h)$ is a generator of $G$.
 A: I will answer the second part of the first question, which should end up encompassing the rest.  The answer is yes, but presentations of groups are slightly subtle, and so the answer is longer than you might expect.
Suppose that a group $G$ has a collection $\{g_{\alpha}\}_{\alpha\in J}$ of generators.  $J$ will be some indexing set, which in nice cases will be finite.  We then have that the elements of the group are represented by words in these generators, and a presentation of the group is the generators, together with some collection of words which are equivalent to the identity.  Let us denote by $R'_G$ the collection of all relations in the generators (in contrast to your $R_G$, which is a given set of relations).  We will adopt similar notations for $H$ as well.
Let $\varphi\colon G\to H$ be a homomorphism (not necessarily an isomorphism yet).  Let $h_{\alpha}=\varphi(g_{\alpha})$.  Because homomrphisms preserve the identity, any relation which holds between the $g_{\alpha}$ must also hold between the $h_{\alpha}$, although we could end up with additional relations.  Moreover, if $h\in H$ is in the image of $\varphi$, then it is of the form $\varphi(g)=\varphi(g_{\alpha_1}g_{\alpha_2}\cdots g_{\alpha_n})=h_{\alpha_1}h_{\alpha_2}\cdots h_{\alpha_n}$.  Therefore, the $h_\alpha$ generate the image of $\varphi$, and they satisfy at least the relations satisfied by the $g_{\alpha}$.
If $\varphi$ is an isomorphism, we can say more.  In particular, the image of $\varphi$ is all of $H$, and so the $h_{\alpha}$ are generators.  Moreover, $\varphi^{-1}$ is also an isomorphism, and so turning the argument around, we see that the relations the $g_{\alpha}$ satisfy are exactly the relations that the $h_{\alpha}$ satisfy.
However, when we have a presentation of a group, we don't have the collection of all relations between the generators, just a small collection of relations.  We have that $R'_G$ is the smallest normal subgroup of $G$ containing $R_G$ (the normal closure).  While we have that $\varphi(R'_G)=R'_H$, we still need to see why $\varphi(R_G)$ and its conjugates actually generates $R'_H$.
However, because every relation in $R'_G$ is a product of conjugates of relations in $R_G$, applying $\varphi^{-1}$ to a relation in $H$, expressing it as such a product, and then applying $\varphi$ to this expression yields that every relation in $R'_H$ comes from relations in $R_H$.
All this is probably more easily stated/proved using the free group  generated by the $h_{\alpha}$, but that is more technology than we really need.
A: Yes and yes. Isomorphisms preserve the group-theoretic structure of a group. And note that if $f: G\rightarrow H$ is an isomorphism then $f^{-1}: H\rightarrow G$ is also an isomorphism, so $2)$ follows immediately from $1)$.
A: Yes, and we can see why in a few different ways.
The easiest is to say that we know that isomorphisms preserve the order of an element. Thus a generator $g$ of $G$ has order $|G|$, and so the order of $f(g)$ is also $|G|=|H|$, and so it generates the whole of that group. The second property is equivalent to the first, as every isomorphism indicates a bijective relationship, and so an isomorphism from $H$ to $G$ also exists.
