For two cyclic groups $\phi:C_4 \to C_3$ with $x^i \mapsto y^i$, we have $\phi(x^4)=\phi(x^0)=e$ but also $\phi(x^4) = y^4=y$. Where am I going wrong?

Question

Let $\phi$ be some function between two cyclic groups of order 4,3 $C_4 =\langle x\rangle $, $C_3 = \langle y\rangle $; $\phi:C_4 \to C_3$, with $\phi(x^i) = y^i$.

Clearly, $x^4 = x^0$, so $\phi(x^4) = y^0 = e$ but by the definition of the function, $\phi(x^4) = y^4 = y$.

My interpretation of the two ways to read $\phi(x^4)$ is clearly leading to a contradiction, and I know the proper reading is $\phi(x^4) = e$, but why isn't $\phi(x^4) = y^4 = y$ implied by the definition? Is it simply a mathematical convention to substitute the input of $\phi$, i.e., $\phi(x^4)$ to $\phi(x^0)$ before evaluating $\phi$?

Answer 1

This shows that $\phi$ cannot be a function. Indeed

$$\begin{align} \phi(x^4)&=y^4\\ &=y \end{align}$$

while

$$\begin{align} \phi(x^4)&=\phi(e)\\ &=\phi(x^0)\\ &=y^0\\ &=e. \end{align}$$

These are two distinct values of $\phi(x^4)$.

Answer 2

Edit: This answers the question before it was edited to assert that $y$ is a generator of target $C_3$.

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This calculation shows that if $\phi$ is going to be a well-defined homomorphism, then we must have $y = e$. In other words, the image of the generator $x$ of $C_4$ has to map to the identity. This is only a contradiction if you previously assumed that, say $y$ was a generator of $C_3$. You don't have to! Just let $y$ be defined as the image $\phi(x)$.

By the way, this shows that any homomorphism $C_4 \to C_3$ must be the trivial one, mapping all elements of $C_4$ to $e \in C_3$.

Answer 3

If $C_n = \langle x \rangle$, and $\phi \colon C_n \to G$ is a group homomorphism, then $$ 1 = \phi(1) = \phi(x^n) = \phi(x)^n, $$ so the order of $\phi(x)$ divides $n$.

This shows that in order to define a group homomorphism with domain a cyclic group of order $n$, we must send a generator of the cyclic group to an element of the codomain whose order divides $n$.

Hence, if we send a generator of $C_4$ to a generator of $C_3$, this does not define a group homomorphism.