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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 $u^{a_1+a_2} v^{2(b_1+b_2)} = u^{a_1}v^{2b_1}u^{a_2}v^{2b_2}$.

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  • $\begingroup$ You need to apply the definition of $M$ to ''exchange'' $u$ and $v$ in the middle term. $\endgroup$ – Wuestenfux Oct 4 '18 at 12:27
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$uv^2u=(uvu)^2=v^2$ so $u$ and $v^2$ commute.

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