Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

So I've asked this type of questions for more than once, and still I don't get the method(s) I've been presented with. What's the general recommended method for finding how many homomorphisms are there, and finding them?

I would probably understand better through an example:

How many homomorphisms are there from $S_3$ to $\mathbb Z_2 \times \mathbb Z_2$.

I would gladly appreciate full solutions.

You have my full gratitude for any sort of assistance, comment, insight or info you can provide.

share|cite|improve this question
I don't think there is an efficient general method. But you can consider all possible maps defined on the generators. Then check which of these induces a well-defined homomorphism. – Seth May 4 '14 at 15:57
up vote 2 down vote accepted

As far as I know, finding a "general method" is quite difficult. Instead, we usually keep some algebraic facts in mind and try to use them to determine some or all of the homomorphisms from one object to another.

In the case of finite groups, one important fact is that if $\phi: G \to H$ is a homomorphism and $g \in G$ has order $n$, then $\phi(g)$ has order dividing $n$. Another important point is that if $g_1, g_2, \ldots, g_k$ are generators for $G$, then $\phi$ is completely determined by where it sends each of the $g_i$'s.

In your example, $S_3$ is generated by $(12)$ and $(123)$. Thus, any $\phi : S_3 \to \mathbb{Z}_2 \times \mathbb{Z}_2$ is determined by picking values for $\phi((12))$ and $\phi((123))$. Since $(123)$ has order 3, $\phi((123))$ must have order dividing 3. The only such element of $\mathbb{Z}_2 \times \mathbb{Z}_2$ is $(0, 0)$, so we must have $\phi((123)) = (0, 0)$. Since $(12)$ has order 2, $\phi((12))$ $\textit{could}$ take any value in $\mathbb{Z}_2 \times \mathbb{Z}_2$. It is then up to you to check that each choice in $\phi((12))$ does indeed yield a group homomorphism. Once you have done this, you have determined all homomorphisms between these groups.

share|cite|improve this answer

Since $\mathbb Z_2\times\mathbb Z_2$ has no nonntrivial elements of odd order, all elements of odd order in $S_3$ must be in the kernel. That is, we need only consider homomorphism from $S_3/Z_3\cong Z_2$. But then we can pick any element of $\mathbb Z_2\times\mathbb Z_2$ as image for the generator (because we only have to make sure that its order divides two). This gives us four homomorphisms $S_3\to\mathbb Z_2\times\mathbb Z_2$.

share|cite|improve this answer

The other answers give correct and reasonably elegant proofs for the special example.

There is a brute force, inelegant method for enumerating group homomorphisms $G \to H$ which will work whenever you have two things:

  • a finite presentation of $G$;
  • a method of calculation in $H$ that lets you determine when two elements are equal.

If you have these two things then here is what you do to enumerate homomorphisms $G \to H$. First, for each generator of $G$, choose its image in $H$; in other words, choose any function from the generating set of $G$ to $H$. Then check that the image in $H$ of each relator of $G$ is equal to the identity in $H$. Enumerating all choices that succeed, you obtain an enumeration of all homomorphisms $G \to H$.

For the special example, the domain $S_3$ has the presentation $\langle a,b \,\, | \,\, a^2=b^2=(ab)^3 = Id\rangle$, and in the range $\mathbb{Z}/2 \times \mathbb{Z}/2$ you can calculate using mod 2 arithmetic. Any map taking $a,b$ to the same element of the range satisfies all three of the relations, and any map taking $a,b$ to different elements of the range satisfies the two order 2 relations but not the order 3 relation. So there are four homomorphisms, each determined by choosing the common image of $a,b$.

share|cite|improve this answer

Let $f:S_3\to Z_2\times Z_2$ be homomorphism;

Since $f$ is a homomorphism, $Ker(f)$ is normal $S_3 $ and normal subgroups of $S_1$ are $\{e\},A_3, S_3$

case 1: if $Ker(f)=\{e\}$ then $f$ is one to one which is nor possible.

case 2: if $Ker(f)=A_3$ then $Im(S_3)\cong S_3/A_3\cong Z_2$ and $Z_2\times Z_2$ has $3$ subgroups which are isomorphic to $Z_2$. Thus, we have $3$ possible $f$.

case 3: if $Ker(f)=S_3$, we have only trivial isomorphism.

Total: We have $4$ different homomorphism.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.