I have some confusion in relation to the following question.
Let $\langle x \rangle = G$, $\langle y \rangle = H$ be finite cyclic groups of order $n$ and $m$ respectively.
Let $f : G \mapsto H$ be the mapping sending $x^i$ to $y^i$.
Determine the conditions on $n$ and $m$ so that $f$ is going to be a homomorphism.
In verifying the definition of a homomorphism I seem to be able to conclude that there is no restriction on $n$ and $m$ since the definition of a homomorphic mapping is satisfied:
$$ f(x^i x^j) = f(x^{i+j}) = y^{i+j} = y^i y^j = f(x^i) f(x^j) \tag{1} $$
But assuming $n < m$ we would get
$$f(x^n) = f(e) = e \not = y^n$$
So clearly we have to have $m \mid n.$
My questions are
- Are there any other conditions on $n$ and $m$ besides $m \mid n.$
- What exactly did I do wrong in $(1)$?
Best Answer
The discussion about "well-defined" is perhaps a bit obscure. Here's the problem:
Remember that in general, an element of $G$ may have many different "names." For example, if $n=10$, then the element $x^{11}$ is equal to $x$; in fact, $x$ has infinitely many different "names" as a power of $x$: $$\cdots = x^{-9} = x^1 = x^{11} = x^{21} = \cdots$$
The problem is that the definition of $f$ as given depends on the name of the element! That is, if we are furiously working and somebody hands us a power of $x$, say, $x^{3781}$, we are supposed to, unthinkingly, map it to $y^{3781}$. The problem is that $x^{3781}$ is the same element as $x$, which we are supposed to send to $y^1$. That means that unless $y^1=y^{3781}$, what we have is not really a function: because the same input, $x$ (who, when being teased by bullies is called "$x^{3781}$") may be sent to $y^1$ or to $y^{3781}$, depending on what "name" we just heard for it.
Checking that the value of the function is the same regardless of what name we are using for an element, even though the function is defined in terms of the name, is called "checking that the function is 'well-defined.'"
An example of a function that is not well-defined would be one in which the input is an integer, and the output is the number of symbols used to express that integer. For example, $f(3)$ would be $1$ (because
3is only one symbol), but $f(4-1)$ would be $3$ (because we are using4,-, and1). This is not well defined as a function of the integers, because $3$ is the same as $4-1$, but $f$ assigns it two different outputs.So in order for the expression given to actually define a function, we need the following to be true: $$\text{if }x^i=x^j,\text{ then }y^i = f(x^i)\text{ is equal to }y^j=f(x^j).$$ Now, $x^i=x^j$ if and only if $i\equiv j\pmod{n}$; and $y^i=y^j$ if and only if $i\equiv j\pmod{m}$. Therefore, we need: $$\text{if }i\equiv j\pmod{m},\text{ then }i\equiv j\pmod{n}.$$ In other words: every number that is a multiple of $m$ must be a multiple of $n$.
This is equivalent to $n|m$.
In fact, you noticed that in what you wrote, because $x^n=e$, so we need $f(x^n)$ to be the same as $f(x^0)$.
Once we know that $n|m$, then $f$ is well defined. Once it is well-defined, we can start verifying that it is indeed a homomorphism (it is). Technically, it's incorrect to start working to see if it is a homomorphism before you even know whether or not it is a function.
Note that the condition actually works if we allow $G$ or $H$ to be infinite cyclic groups, if we call infinite cyclic groups "groups of order $0$". Then $i\equiv j\pmod{0}$ means $i=j$, every number divides $0$, but the only multiple of $0$ is $0$. So if $G$ is infinite cyclic then the value of $n$ does not matter; if $H$ is infinite cyclic then we must have $G$ infinite cyclic.