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I started self-learning algebra and am proving isomorphisms. I think of one approach which I tried to formalize as follows:

Given group $G=\langle S\mid R\rangle$ where $S=\{a_1,\ldots, a_n\}$ and $R = \{r_i(1,a_1,\ldots,a_n)\}_{i=1}^m$. If we have another group $G'$ and function $\varphi:G\rightarrow G'$ such that $\varphi(1_G)=\varphi(1_{G'})$ and $r_i(1_{G'},\varphi(a_1),\ldots,\varphi(a_n))$ hold for $i=1,\ldots, m$, where the law of composition in $r_i$s are replaced by that of $G'$. Then $G\approx{\rm Im}(\varphi)$.

If this holds, then to prove $G\approx G'$ I only need to consider the map between their generators and check the relation of $G$ hold for $G'$ w.r.t. the corresponding generators. However, I cannot prove or disprove the statement above. I think the problem lies in how to formalize the group presentation and its properties.

Inspired by the comments: What if $\varphi$ is defined from $S\rightarrow G'$? Can we extend it to a homomorphism from $G\rightarrow G'$ given that $\varphi(a_1),\ldots,\varphi(a_n)$ follow the same rules in $R$? (This seems to be proved in Lee's answer in problem When can a homomorphism be determined entirely by its generators)

LonelyQuantum
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    This is not true. You have to show the only relations between the generators in $G’$ are the ones derived from the $r_i.$ (Also, if $f$ is really just a function, and not a homomorphism, then your condition doesn’t require $f$ to be a homomorphism.) – Thomas Andrews Aug 18 '21 at 16:45
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    @ThomasAndrews Maybe I'm missreading something (in particular, there is no $f$ while the $f_i$s are not defined...) but surely what this set-up gives you is precisely that the map $\varphi$ is a homomorphism? The set up says that the relations of the group satisfy the relations of the codomain-group, which is what we need to get a homomorphism. – user1729 Aug 18 '21 at 18:09
  • If you are using relations loosely, maybe, but usually, the relations in the notation $\langle S\mid R\rangle$ notation of a group are things like: $$a_1^2=1,a_2^n=1,a_1a_2a_1a_2=1$$ @user1729 This defines the dihedral group. But the relations are just three conditions on the generators. – Thomas Andrews Aug 18 '21 at 18:12
  • @ThomasAndrews But in this question it is clear that $R$ is intended to be a set of defining relators rather than relations, i.e. words in the generators that evaluate to the identity in the group being defined. There is nothing loose about that - that is the way that group presentations are formally defined, and the group defined is the quotient of the free group by the normal closure of the defining relators. – Derek Holt Aug 18 '21 at 18:24
  • That is far from obvious in the way the question is written to me, @DerekHolt In particular, noting says there are finite elements in $G,$ but there are finite relations. – Thomas Andrews Aug 18 '21 at 18:31
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    @ThomasAndrews ah, wait, I understand: the point is that the function is from $G\rightarrow G'$. If it were from $S\rightarrow G'$ and the images of the relators were trivial in $G'$ then it would extend to a homomorphism. (Unless the statement has something additional about multiplication being respected, but maybe that is the bit with the $f_i$s, and I'm struggling to parse that bit!) – user1729 Aug 18 '21 at 18:31
  • (I agree with @Derek's interpretation of the question - this is precisely what the notation $\langle S\mid R\rangle$ means, and so this meaning is clear from the question.) – user1729 Aug 18 '21 at 18:35
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    Yes, I see the problem, the given function should be from $S$ to $G'$ , and we need to prove that it can be extended to a homomorphism from $G$. – Derek Holt Aug 18 '21 at 18:36
  • Could you please explain why I can extend the map from $S$ to $G'$ @DerekHolt to a homomorphism from $G$ to $G'$. I think that is exactly what I tried to do and the formalization in the question has some problems. – LonelyQuantum Aug 18 '21 at 18:45
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    The condition that $r_i(1_{G'},\varphi(a_1),\ldots,\varphi(a_n))$ is equal to $1_{G'}$ for $i=1,\ldots, m$ guarantees that the map $\phi:S \to G'$ extends to a homomorphism $\varphi:G \to G'$. That is a basic result in the theory of group presentations. – Derek Holt Aug 18 '21 at 19:05
  • @DerekHolt Thank you. I think I finally read about it in Chpater 7 of Artin's book. It is an extension of the First Isomorphism Theorem. – LonelyQuantum Sep 12 '21 at 16:21

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If $G'$ is also defined by a finite presentation (or if a finite presentation for $G'$ can be easily computed), then a good way to prove that $\varphi$ is an isomorphism is to find its inverse.

To do that, you need to find a homomorphism $\varphi':G' \to G$, using the same technique as for $\varphi$, and then to verify that the composites $\varphi'\circ \varphi:G \to G$ and $\varphi\circ \varphi':G \to G'$ are both equal to the identity map.

Derek Holt
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