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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?

Frunobulax
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spectraa
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    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. – angryavian Sep 08 '14 at 11:58
  • Are you asking "when does every choice of generator produce a homomorphism"? This is different from your states question (but is more interesting!). – user1729 Sep 08 '14 at 16:04
  • @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. – spectraa Sep 08 '14 at 16:09
  • So does my answer help or not? If not, I'll delete it. – user1729 Sep 08 '14 at 16:09
  • @user1729 that means that we'll put such a restriction on generators such that it is satisfied to be a homomorphism always.that seems good theoretically but how can we think this way while constructing some example .Please help.. – spectraa Sep 08 '14 at 16:13
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    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 $r_j$. – Lee Mosher Sep 08 '14 at 16:39
  • Hi @Lee Mosher, do you have a reference in a text where this statement can be found? Thanks – user50229 Aug 19 '15 at 10:20
  • @user50229: I'm not sure of a reference, I would guess it is in any of the older combinatorial group theory books such as Magnas, Karass, and Solitar. But I'll add another answer with a sketch of the proof. – Lee Mosher Aug 19 '15 at 11:58

3 Answers3

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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 $f$.

Lee Mosher
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  • +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. – yoyostein Apr 28 '18 at 09:23
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    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. – Lee Mosher Apr 28 '18 at 15:03
  • 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. – yoyostein May 02 '18 at 15:47
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    Each $r_j$ is a word in the letters $g_i$, say $r_j = g_{i_1} ... g_{i_m}$, so $f(r_j) = f(g_{i_1}) ... f(g_{i_m})$. – Lee Mosher May 02 '18 at 20:22
  • @LeeMosher I thought each $r_i$ is an equation, such as $rs = sr^{-1}$. In formal presentations, do we actually make one side the identity in each equation, such as $rs(sr^{-1})^{-1}=1$, and then just write the word $rs(sr^{-1})^{-1}$ instead of the equation? – Blue Nov 23 '19 at 18:54
  • That's correct. Formally, that word is interpreted as an element of the free group on the generating set. Take a look at the formal definition of a group presentation. – Lee Mosher Nov 23 '19 at 21:33
  • @LeeMosher Thanks! – Blue Nov 23 '19 at 21:46
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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!

Community
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user1729
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    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 $G$. – user1729 Sep 08 '14 at 16:29
  • 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. – spectraa Sep 08 '14 at 16:32
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    @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. – user1729 Sep 08 '14 at 19:34
  • it was helpful. thanks once again. – spectraa Sep 09 '14 at 02:25
  • 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 ? – spectraa Sep 09 '14 at 04:26
  • edit to my previous comment G is abelian. – spectraa Sep 09 '14 at 04:35
  • @WantTobeAbstract Your map is for all group elements, not just the generators! If it was for just the generators, then you first need a presentation of your group $G$. The relations are the relations of this presentation. For example, If $G$ was the group $\mathbb{Z}_6\times\mathbb{Z}_2=\langle a, b; ab=ba, a^6=1, b^2=1\rangle$ and $n=2$ then you need to verify that $a^2b^2=b^2a^2$, that $a^{12}=1$ and that $b^4=1$. These three relations all clearly hold, so the map $a\mapsto a^2$, $b\mapsto b^2$ is a homomorphism. – user1729 Sep 09 '14 at 08:15
  • sorry to disturb you again with doubt on above post:I understood about checking homomorphism whenever presentation of group is known.But I had a question in my exam which said :Suppose $\phi$ is a ring homomorphism from $Z\oplus Z$ into $Z\oplus Z$.What are possibilities for $\phi ((1,0))$?I had the idea to search possibilities using presentation of ring. I didn't know what presentation to be made for ring,I just wrote the group presentation of it under $+$ ,and under multiplicationabelian.I still don't know how to write presentation of any ring.Please help how I should do it.... – spectraa Oct 01 '14 at 03:33
  • @spectraa You should ask this as a new question. – user1729 Oct 04 '14 at 17:03
  • I had asked a similar question on writing the presentation of ring but it proved of no help to me as the answers describe it something to do with free groups which I've not done yet .The link to that question is : http://math.stackexchange.com/questions/949057/checking-whether-a-map-satisfies-being-homomorphism – spectraa Oct 04 '14 at 17:07
  • @spectraa How are you learning abstract algebra? Free objects are required for the definition of a presentation. For example, you cannot give a presentation for a field, as there is no such thing as a "free field". So, in general, free groups come before presentations. – user1729 Oct 05 '14 at 16:38
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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 $aba^{-1} = b^2$.

Martin Brandenburg
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