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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:

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$$ f(x^i x^j) = f(x^{i+j}) = y^{i+j} = y^i y^j = f(x^i) f(x^j) \tag{1} $$

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 $(1)$?

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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 `3`

is only one symbol), but $f(4-1)$ would be $3$ (because we are using `4`

, `-`

, and `1`

). 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.

You may want to check first if the map is well defined, ie., if $x^i=x^j$ does it follow that $f(x^i)=f(x^j)$?

I am not sure if this completely explains why the naive calculations in $(1)$ are not valid. Here is my go at it:

I’ll list down several points of concern.

- Firstly your map $x^i \mapsto y^i$ is very naive that it misses some essential details.

For convenience of notation, I prefer to look at this as $\varphi(\bar i)=\bar i$ if you’ll allow me to do so. This is also naive but to some extent captures what is happening.

So, an equivalence class $\bmod n$ is mapped to the same equivalence class $\bmod m$ under $\varphi$.This bit of information is missed in $x^i \mapsto y^i$. This explains my comment under your question that you will want to write your maps as $x^{i(n)} \mapsto y^{i(m)}$. This notation makes things more transparent.

That you have dicovered that $m \mid n$, it can be obtained through the universal property of the quotients, which I will write about later.

The quotient map $G \to G/H$ factors over $G \to G/K$ if and only if $K \subseteq H$. That is there is a map from $G/H$ to $G/K$ if and only if $K \subseteq H$.

**Proof**: Please Go through Dummit and Foote’s description in their Abstract Algebra in the section $\S3.3$ *Isomorphism Theorems.*

Now consider your cyclic groups as $\Bbb Z/n\Bbb Z$ and $\Bbb Z/m\Bbb Z$ and observe that above condition will tell you, $m\Bbb Z \subseteq n \Bbb Z$ whic happens if and only if…

As you had earlier in your comments observed, the map won’t even be well defined if $m \nmid n$.

I’ll leave it to you as a challenge to use this blooper to prove that two groups of same order are isomorphic, which really is not the case. **Hint**: Cyclic groups exist for any given order.

The rule $x^i\mapsto y^i$ gives a perfectly well-defined function in the set $\{x^0, x^1, \ldots, x^{n-1}\}$. However, for that rule to be a homomorphism we need that $x^{i+j} \mapsto y^{i+j}$ for all $i$ and $j$. To find where $x^{i+j}$ goes we need to reduce $i+j \bmod n$ to a $k$ in $\{0,1,\ldots,n-1\}$ and we need that $k \equiv i+j \bmod m$. In other words, we need $k \equiv k' \bmod n \implies k \equiv k' \bmod m$ and this can only happen when $m\mid n$.

THEOREM:

Define $f:\langle x \rangle \to \langle y \rangle$,

where the orders of

$\langle x \rangle$ and $\langle y \rangle$

are n and m respectively,

by $f(x) = y^u$. Then f is a well-defined homomorphism if and only if

$\frac{m}{\gcd(m,n)} \mid u$.

PROOF:

Suppose that the mapping

$f:\langle x \rangle \to \langle y \rangle$

is a well-defined homomorphism.

Let

$f(x) = y^u$ for some non negative integer u.

We must also have

$ 1 = f(x^n) = y^{un}$

Then

$ m \mid un$.

Let

$g = \gcd(m, n)$.

Then

$\frac m g \mid u \frac n g$.

It follows that

$\frac m g \mid u$.

The converse is pretty straightforward.

Suppose we define $f:\langle x \rangle \to \langle y \rangle$

by $f(x) = y^u$ where $\frac{m}{\gcd(m,n)} \mid u$

The only real problem with this definition is “Is it well defined?”

Since the order of $\langle x \rangle$ is $n$:

\begin{align*}

x^a = x^b &\Rightarrow a \equiv b \mod n \cr

&\Rightarrow ua \equiv ub \mod {un} \cr

&\Rightarrow ua \equiv ub \mod {\frac{mn}{\gcd(m,n)}} \cr

&\Rightarrow ua \equiv ub \mod {m} \cr

&\Rightarrow y^{ua} = y^{ub}\cr

&\Rightarrow f(x^a) = f(x^b)

\end{align*}

Linearity is pretty obvious, so $f$ is a well-defined homomorphism.

So, if you want $f(x^i) = y^i$, then you want $f(x) = y$.

You must therefore have $\frac{m}{\gcd(m,n)} \mid 1$. Which implies $m \mid n$.

Conversely, if $m \mid n$ and $f(x) = y$, it follows that $f(x^i) = y^i$.

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