Homomorphism between cyclic groups of order 4 and 2

Question

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!

Answer 1

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 $\varphi (x^p)=\varphi(x^q)$.

Answer 2

Or, you could do an inclusion homomorphism, since $\mathbb Z_4$ contains a copy of $\mathbb Z_2$ as a subgroup. Actually, that's the same one as $h (x)=2x $. There's only one nontrivial way to do it, because a homomorphism out of a cyclic group is determined by where it sends $1$, say. And the only nontrivial place to send it is to $2$, since the order of $h (1) $ must divide $2$.