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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 $G$.

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Is this a homework question? –  Asaf Karagila May 1 '11 at 6:40
How are you defining a generator of a group? It is more usual to refer to a generating set. The second part of your first question is more in that vein. In any case the answer will be yes and yes! (as was said below). –  Tim Porter May 1 '11 at 9:23

3 Answers 3

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.

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I think that this is the correct approach. Seeing that groups are the quotients of the free group really clears up group presentations and prepares one for other algebraic structures. –  BBischof May 1 '11 at 15:41
@BBischof I agree, but it takes some work to show that free groups exist, that they satisfy their universal property, that they are described the way you want them to be described, that they related to presentations the way you want them to relate to presentations, etc. Conceptually they are a boon, and it is easy to believe that all the basic true things about them are true, but it is a bit far afield for anybody just learning groups theory. –  Aaron May 1 '11 at 16:26
my personal experience was that I learned them smack dab in the middle of my group theory course, just before rings. I found them to be super intuitive, and they made me realize I was not understanding some things about groups that I though I had. –  BBischof May 1 '11 at 16:51
@BBischof In that case, I applaud your professor for introducing infinite, non-abelian groups which weren't the classical Lie groups. I was under the impression that most introductory books didn't introduce groups that they didn't need for either rep theory or the structure theorem for f.g. modules over PIDs. Free groups are covered almost as an afterthought in Dummit and Foote, and I don't think they are covered at all in Artin. It is probably in Lang, but so are spectral sequences, so I don't know what a course out of Lang would look like. –  Aaron May 1 '11 at 17:42
I should mention that this was not in our actual course. It was supplementary lectures. I can envision a nice course where free groups come fairly naturally. Either based on D&F or just in general. –  BBischof May 1 '11 at 19:30

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 $1)$.

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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.

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What about infinite groups? For example $2\in \mathbb{Z}$ has infinite order but does not generate $\mathbb{Z}$. –  Alex Becker May 1 '11 at 6:27
@Alex: This is a good point, I had assumed this to be a simple question, and therefore limited it to only finite groups. I chose the simplest concrete idea I had for why isomorphisms preserve structure. –  mixedmath May 1 '11 at 6:36

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