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.MostlyHarmless said: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?
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}}}##.MostlyHarmless said: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.
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.MostlyHarmless said: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.