Why is this homomorphism?

Let $$M=\langle u,v: u^2=v^8=1, vu=uv^5\rangle$$ be a group of order $$16$$.

I have a reading, that says:

Let $$Z_2=\langle x\rangle$$ and $$Z_4=\langle y\rangle$$, then $$\varphi:Z_2\times Z_4\to M$$ given by $$(x^a,y^b)\to u^av^{2b}$$ is an injective homomorphism.

Then they go on to show that $$\varphi$$ is indeed well-defined, and is injective, but I have real problem with their homomorphism proof: It goes like

Let $$(x^{a_1},y^{b_1}), (x^{a_2},y^{b_2}) \in Z_2 \times Z_4.$$ Then $$\varphi((x^{a_1},y^{b_1}) (x^{a_2},y^{b_2})) = \varphi(x^{a_1+a_2}, y^{b_1+b_2}) = u^{a_1+a_2} v^{2(b_1+b_2)} = u^{a_1}v^{2b_1}u^{a_2}v^{2b_2} = \varphi(x^{a_1},y^{b_1}) \varphi(x^{a_2},y^{b_2})$$

I can't get how do we get $$u^{a_1+a_2} v^{2(b_1+b_2)} = u^{a_1}v^{2b_1}u^{a_2}v^{2b_2}$$.

• You need to apply the definition of $M$ to ''exchange'' $u$ and $v$ in the middle term. – Wuestenfux Oct 4 '18 at 12:27

$$uv^2u=(uvu)^2=v^2$$ so $$u$$ and $$v^2$$ commute.