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Consider the following multiple choice problem:

Let $H$ be the set of all group homomorphsims $\phi:{\bf Z}_3\to{\bf Z}_6$. How many functions does $H$ contain?

A.1 B.2 C.3 D.4 E.6

Since $1$ generates ${\bf Z}_3$, one can analyze $\phi(1)$ case by case, which may be rather time consuming, for me, at least. Since this is a multiple choice problem, is there any quick way to solve it?

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5 Answers5

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  1. A group homomorphism with cyclic domain is completely determined by the image of a generator.

  2. If $f\colon G\to H$ is a homomorphism, and $x\in G$, then the order of $f(x)$ must be a divisor of the order of $x$.

Since the only divisors of $3$ are $1$ and $3$, the answer is one plus the number of elements of $\mathbb{Z}_6$ of order $3$ (the "one plus" comes from the trivial map). Are there any? Yes, so the answer is not A. How many? Two: 2 and 4. So the answer is C.

Arturo Magidin
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  • Now I see. Thanks, Arturo. Without 2, I might do it very slowly. –  Jun 16 '11 at 04:23
  • @Jack: And you do see why it is true, right? – Arturo Magidin Jun 16 '11 at 04:24
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    I guess you are using the fact that the order of $x$ is equal to the order of the cyclic group $$. Since the order of $<f(x)>$ divides the order of $f()$, and the order of $f()$ divides the order of $$, then '2' follows. Correct? –  Jun 16 '11 at 04:30
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    @Jack: That works, but simpler: if $x^n=1$, then $f(x)^n = f(x^n) = f(1) = 1$, so the order of $f(x)$ must divide $n$. In particular, if $n$ is the order of $x$, you get 2. – Arturo Magidin Jun 16 '11 at 04:31
  • I don't understand 1. What does it mean when you say homomorphism is completely determined? And second question, I thought $|f(a)| $ divides $ |a|$ is necessary condition for $f$ to be homomorphism. So how does choosing $f(a)$ such that it's a divisor of $|a|$ ensure $f$ is homomorphism? – William Dec 28 '21 at 15:18
  • @William: "Completely determined" if you know what happens to a generator, you know what happens to everything. Second question: because the condition you give is for arbitrary groups (as I note in point 2), but here we have a special situation in which the domain is a cyclic group. In that situation, the condition suffices to define a homomorphism when $a$ is a generator. Work it out. – Arturo Magidin Dec 28 '21 at 15:31
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In general the number of group homomorphisms $\varphi:\mathbb{Z}_{m} \to \mathbb{Z}_{n}$ is given by $\text{gcd}(m,n)$. So here you have $\text{gcd}(3,6)=3$.

The proof of this result can be found in Abstract Algebra Manual: Problems and Solutions By Ayman Badawi.

  • I like the "general" one. Knowing such theorem is necessary for do it "very quick". –  Jun 16 '11 at 04:24
  • @Jack: Yes, true –  Jun 16 '11 at 04:24
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    Any reference for this proposition? I did learn this when I learned the basic group theory. –  Jun 16 '11 at 04:33
  • @Jack: Wait for sometime, i shall add it. –  Jun 16 '11 at 04:40
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    For anyone wondering what is a ring function such as $\rm gcd$ doing in group theory: it might be useful to recall that every abelian group is actually a $\mathbb Z$-module. – Marek Jun 16 '11 at 21:16
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I just learned that there are two general theorems about this question

  1. The number of group homomorphisms from $Z_m$ into $Z_n$ is $\text{gcd}(m,n)$.
  2. The number of ring homomorphisms from $Z_m$ into $Z_n$ is $2^{\omega(n)-\omega(n/\text{gcd}(m,n))}$ where $\omega(a)$ denotes the number of distinct prime divisors of the integer $a$.

from an article The Number of Homomorphisms from $Z_m$ into $Z_n$(American mathematical Monthly 91 (1984):196-197) by Gallian and Buskirk.

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Say $f$ be the homomorphism and say $f(1)=a\in \mathbb{Z}_6$

in $\mathbb{Z}_3$ we have $1+1+1=3=0\Rightarrow f(1+1+1)=f(0)=0\Rightarrow 3f(1)=0\Rightarrow 3a=0$ in $\mathbb{Z}_6$ so $a=0,2,4$

so you have $3$ distinct homomorphism.

Myshkin
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Order of 1 is 3 order of image of 1 divides 3 and #(f(1))/6 so so total choice is 1 and 3. Here there are there are two choice of order 3 element so total 3 homomorphism.

Davide Giraudo
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