I need to show that there is an homomorphism between cyclic groups of order 4 and 2. Its the first time I'm studying group theory so I'm fairly confused about what i need to do here. My attempt was the following:
I need to show that there is a map $\varphi : \mathcal{G} \to \mathcal{H}$ that preserves the group multiplication, where $\mathcal{G}$ is a cyclic group of order 2 and $\mathcal{H} $ a cyclic group of order 4. So i attempet to define the map as follows:
$$
\mathcal{G}=\{ x^0=e, x^1\}; \space\mathcal{H}=\{y^0 , y^1, y^2,y^3\}\\
\varphi:\mathcal{G}\to\mathcal{H}\\
\varphi(x^n)=y^{2n}
$$
Witch preserves group multiplication since:
$$
\varphi(x^nx^m)=\varphi(x^{n+m})=y^{2(n+m)}=y^{2n}y^{2m}=\varphi(x^n)\varphi(x^m)
$$
Honestly i doubt this makes much sense so i decided to ask about it here. Any clarifications are much appreciated!
Best Answer
Note that because elements in finite cyclic group have more than one way to be expressed (eg. In $\mathcal{G}$, $x^5=x^1$). You have to show that the map is well-defined, that is, for integers $p,q$, $x^p=x^q$ implies that $\varphi (x^p)=\varphi(x^q)$. This can be proven as follows:
Since $x^p=x^q$, we have $x^{p-q}=1$ and therefore $2$ divides $p-q$. This implies that $4$ divides $2(p-q)=2p-2q$. It follows that $y^{2p-2q}=1$, i.e., $y^{2p}=y^{2q}$ and therefore we prove that $\varphi (x^p)=\varphi(x^q)$.