A group homomorphism that doesn't send one into one.

Question

Let $G$ be a group of order 3, $G=\langle g\rangle$ and let $H$ be a group with an element h of order 4. Let $f: G\longrightarrow H$, where $f(g^i)= h^i$. Then f is not an homomorphism because $f(g^3)= h^3$ but $g^3= 1$ and $h^3 \neq 1$. However

$$\begin{align} f(g^ig^j)&= f(g^{i+j})\\ &= h^{i+j}\\ &= h^i h^j\\ &= f(g^i)f(g^j). \end{align}$$

Where is the mistake?

In particular $f(gg^2)=f(g^3)=h^3=hh^2=f(g)f(g^2)$.

Answer 1

That map $f$ doesn't make sense. Note that $g^4=g$, but $f(g^4)\neq f(g)$.

Answer 2

Let $G=\lbrace 1, g, g^2 \rbrace$ and $H$={$e, h, h^2, h^3 $}, where $ e$ is the identity of $H$. Define $f$ as follows: $f:G\rightarrow H$ such that $f(g^i)=h^i$ that is, $f(1)=e$, $f(g)=h$, $f(g^2)=h^2$ and $h^3$ is not an image of $G$. Consequently, $f(g^2g)=f(g^3)=f(1)=e=h^4=h^3 h=h^3f(g).$

Hence $f$ is not a homomorphism (i.e., when $h^3\neq h^2=f(g^2)$). And in this case $f(g^3)=h^3$ is wrongly defined.