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I read a text which says that:

Just because a homomorphism $ϕ :G → H$ is determined by the image of its generators does not mean that any such image will work.

e.g.: Suppose we try to define homomorphism $ϕ :Z_3 → Z_4$ by $ϕ(1)=1$ , then we get $ϕ(0)=ϕ(1+1+1)=3$ which isn't possible as $ϕ(0)=0$.

Does there exist some case in which a homomorphism is entirely determined by its generators?

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    $\begingroup$ Say $G$ has generators that satisfy a set of relations (e.g., $g+g+g=0$ in your example). If $\phi$ sends the generators of $G$ to elements in the image in a way that satisfies all these relations, then you know that $\phi$ is a homomorphism. $\endgroup$ – angryavian Sep 8 '14 at 11:58
  • $\begingroup$ Are you asking "when does every choice of generator produce a homomorphism"? This is different from your states question (but is more interesting!). $\endgroup$ – user1729 Sep 8 '14 at 16:04
  • $\begingroup$ @user1729 Yes I had both doubts 1.)when does every choice of generator produce a homomorphism 2.) a homomorphism is entirely determined by its generators.your answer made 2.) clear. $\endgroup$ – spectraa Sep 8 '14 at 16:09
  • $\begingroup$ So does my answer help or not? If not, I'll delete it. $\endgroup$ – user1729 Sep 8 '14 at 16:09
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    $\begingroup$ The comment of @angryavian deserves greater emphasis, because it it answers one version of your question which may be what you actually want to know. Namely, if you have a presentation $\langle a_i \, | \, r_j \rangle$ of the group $G$, and if you are given values of $\phi(a_i) \in H$ for all generators $a_i$, then this extends to a homomorphism $\phi : G \to H$ if and only if $\phi(r_j)$ is the identity element of $H$ for each relator $r_j$. $\endgroup$ – Lee Mosher Sep 8 '14 at 16:39
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A homomorphism is always determined by its generators, whether it is an isomorphism or not. To be explicit:

Q: does there exist some case in which a homomorphism is entirely determined by its generators?

A: Yes, every single possible case. A homomorphism is always defined by its generators.

All the example is saying is that you cannot just take some map of the generators and hope that it is a homomorphism.

Three more examples:

  • Any map $\phi: \mathbb{Z}_n\rightarrow\mathbb{Z}$, $1\mapsto x$, for $x\neq 0$, is not a homomorphism as if it is then $n\cdot x=0$, a contradiction!

  • Suppose $\gcd(n, m)=1$. Then $\phi: \mathbb{Z}_m\rightarrow\mathbb{Z}_n$, $1\mapsto x$, for $x\neq 0\pmod n$, is not a homomorphism because the image of the subgroup $\mathbb{Z_m}$ must have order dividing $m$ (why?) but all subgroups of $\mathbb{Z}_n$ have order dividing $n$. (This is a generalisation of the example given in your question.)

  • Suppose $G$ is simple and $H$ contains no subgroup isomorphic to $G$. Then any map $\phi: G\rightarrow H$ where the generators are not mapped to the identity of $H$ is not a homomorphism. For example, the generators in a map $\phi: A_5\rightarrow \mathbb{Z}_n$ must be sent to the identity, otherwise the map is not a homomorphism.

Of course, you may be asking "when does every choice of generator produce a homomorphism". If that is so, this is not clear. But then you should read Martin's answer!

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  • $\begingroup$ I am unsure precisely what you are asking, but I think the restriction you are after is if your map is $\phi: G\rightarrow H$ then you want the generators to generate a subgroup of $H$ which is isomorphic to a homomorphism of $G$. In my first example, $\mathbb{Z}$ contains no elements of finite order but all homomorphic images of $\mathbb{Z}_n$ are finite cyclic. In the third example, $G$ has only itself and the trivial group as a homomorphic image, but $H$ does not contain a copy of $G$. $\endgroup$ – user1729 Sep 8 '14 at 16:29
  • $\begingroup$ to be more clear I mean to say that suppose I've to construct a homomorphism $\phi :$$G$$\rightarrow$$H$,to define a mapping on the generator then I'll have to check it whether it works for every element of group G.Wouldn't it be tedious.Can you help me by illustrating it through an example. $\endgroup$ – spectraa Sep 8 '14 at 16:32
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    $\begingroup$ @WantTobeAbstract Ah, okay, you just need to check that the homomorphism works for the relators of a given presentation (with your given generators). You might find this answer of mine useful. $\endgroup$ – user1729 Sep 8 '14 at 19:34
  • $\begingroup$ it was helpful. thanks once again. $\endgroup$ – spectraa Sep 9 '14 at 2:25
  • $\begingroup$ I've just come across a homomorphism $\phi : G \rightarrow G$ by $\phi(g)=g^n$ .(in additive group we replace $g^n$ by $ng$. Now while constructing this homomorphism what relations must have been satisfied.I feel there's just one that identity must map to identity element.Are there any other relations that must be satisfied also here ? $\endgroup$ – spectraa Sep 9 '14 at 4:26
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I will expand my comment into an answer:

Theorem: Given a group with presentation $$G = \langle \, g_i \bigm| r_j \, \rangle $$ given another group $H$, and given a function $f$ which assigns to each $g_i$ a value $f(g_i) \in H$, the function $f$ extends to a homomorphism $F:G \to H$ if and only if each $f(r_j)$ is the identity.

Proof: Let $\langle g_i \rangle$ denote the free group with free basis $\{g_i\}$. Let $N$ be the smallest normal subgroup of $\langle g_i \rangle$ containing $\{r_j\}$. By definition of presentation, $$G = \langle g_i \rangle \, / \, N $$ Let $q : \langle g_i \rangle \to G$ be the quotient homomorphism.

By the universal property for free groups, the function $f$ extends to a homomorphism $\widetilde F : \langle g_i \rangle \to H$. Let $K = \text{kernel}(\widetilde F)$. Then we have the following chain of equivalences: $\{r_j\} \subset K$ $\iff$ $N < K$ $\iff$ there is a homomorphism $F : G \to H$ such that $F \circ q = \widetilde F$ $\iff$ there is a homomorphism $F : G \to H$ that extends $f$.

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  • $\begingroup$ +1. I think your answer is very good and completely answers the question in full generality. Is this found in any book? I would like to read up further on related topics. $\endgroup$ – yoyostein Apr 28 '18 at 9:23
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    $\begingroup$ Thank you. I'm not sure what book to recommend, although the topic is covered in books on combinatorial group theory. One of the older books is Magnus/Karass/Solitar, although it sometimes does not adopt the cleanest abstract point of view. $\endgroup$ – Lee Mosher Apr 28 '18 at 15:03
  • $\begingroup$ Just to clarify something. If $f$ is only defined on each $g_i$, how do we know, a priori, what is $f(r_j)$? Thanks a lot. $\endgroup$ – yoyostein May 2 '18 at 15:47
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    $\begingroup$ Each $r_j$ is a word in the letters $g_i$, say $r_j = g_{i_1} ... g_{i_m}$, so $f(r_j) = f(g_{i_1}) ... f(g_{i_m})$. $\endgroup$ – Lee Mosher May 2 '18 at 20:22
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By the fundamental theorem on homomorphisms, $\hom(\mathbb{Z}/n\mathbb{Z},G) \cong \{g \in G : g^n=1\}$, the $n$-torsion of $G$. In other words, $g^n=1$ is the only relation which is required by the image $g$ of the canonical generator of $\mathbb{Z}/n\mathbb{Z}$. One can easily derive from this $\hom(\mathbb{Z}/n\mathbb{Z},\mathbb{Z}/m\mathbb{Z}) \cong \mathbb{Z}/\mathrm{ggT}(n,m)$.

If $E$ generates $G$, then $\hom(G,H) \to \mathrm{Map}(E,H)$ is injective. This is what one means by "a homomorphism is determined by the images of the generators". It is surjective (for all $H$) if and only if $E$ is a free generating set of $G$, i.e. $G= F(E)$ is a free group. Only in this case, every choice of the images produces a homomorphism.

In general, a group presentation exactly contains the information about the relations which are necessary for defining a homomorphism. For example, $G=\langle x,y : x^2 = y^5=1 , xyx^{-1} = y^2 \rangle$ is the group with the property that homomorphisms $G \to H$ correspond to elements $a,b \in H$ (the images of $x,y$) such that $a^2=b^5=1$ and $aba^{-1} = b^2$.

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    $\begingroup$ "ggT" or "größte gemeinsame Teiler" mean gcd in German. $\endgroup$ – Orat Apr 2 '18 at 1:48

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