# Group Presentations Isomorphisms

1. Dec 13, 2015

### MostlyHarmless

If you are trying to show that two groups, call them H and G, are isomorphic and you know a presentation for H, is it enough to show that G has the same number of generators and that those generators have the same relations?

Last edited: Dec 13, 2015
2. Dec 13, 2015

### Staff: Mentor

I'd say yes. You can map the first set of generators to the second which gives you a surjective group homomorphism. Its kernel are all relations which map onto the relations in H which come from relations in G, i.e it's injective, too. Maybe one has to be a little more careful if generators or relations are infinitely many, but I can't see what might be wrong. And otherwise it wouldn't make much sense to describe a group that way.

3. Dec 13, 2015

### MostlyHarmless

It's not obvious to me why simply mapping the generators to generators should define a homomorphism.

Just to be sure I'm on the same page as you. Let $a,b$ generate G, $a',b'$ generate H. Let $\phi : G \rightarrow H$ be defined by $\phi(a)=a'$ and $\phi(b)$.

From just this, It's not clear how one would show that $\phi$ defines a homomorphism. Of course I believe it does, and of course saying what a map does to the generators should be enough to describe what it does to the group.

And I agree, showing that two groups have presentations that look the same upto naming generators should be enough to claim isomorphic.. I'm just unclear on how to show this rigorously.

4. Dec 13, 2015

### Staff: Mentor

Each element $g$ of $G$ can be written as a word in the generators $a_i$, say $g = ∏ a_{i_{k}}^{n_{i_{k}}}$ for a finite sequence $(i_k)$. Defining $Φ(g) = ∏ {a'}_{i_{k}}^{n_{i_{k}}}$ sets up a surjective group homomorphism, i.e. $Φ(gh) = Φ(g)Φ(h)$ - just the concatenation of the words in either group. It remains to show that $\{g ∈ G | Φ(g) = e\}$ is the set of relations in $G$. But $Φ(g)=e$ are exactly the relations in $H$ which are those in $G$ without '. Or the other way: One can immediately define $Φ^{-1}$ and therefore the relations have to be the same. I think the only problem is to make sure $Φ$ is well-defined. However $∏ {a}_{i_{k}}^{n_{i_{k}}} = ∏ {b}_{j_{k}}^{m_{j_{k}}}$ defines a relation in $G$ which translates to one in $H$, i.e. $∏ {a'}_{i_{k}}^{n_{i_{k}}} = ∏ {b'}_{j_{k}}^{m_{j_{k}}}$.

I don't see a trap.

5. Dec 13, 2015

### WWGD

Another idea: Given groups G,H , if you can express every h in H as a word in G and every g in G as a word in H, you have shown G,H are isomorphic.