Prove the Universal Mapping Property for Z x Z:
If g and h are any elements of a group G, then there exists a unique homomorphism φ: Z x Z --> G such that φ(1,0)=g and φ(0,1)=h if and only if gh=hg.
Assuming that there exists a unique homomorphism φ, with g,h in G, then gh=φ(1,0)φ(0,1)=φ(1,1) since φ is a homomorphism, then φ(1,1)=φ(0,1)φ(1,0)=hg
I think this is right but I am having trouble proving it the other way. Any help would be much appreciated, thank you!
Define $\phi : \Bbb Z\times\Bbb Z\to G$ by $\phi((n,m)) = g^n h^m$. Then \begin{align*} \phi((n,m))\phi((n',m')) &= g^nh^mg^{n'}h^{m'}\\ &= g^n g^{n'}h^m h^{m'}\\ &= g^{n+n'} h^{m+m'}\\ &= \phi((n,m) + (n',m')) \end{align*} because $g$ and $h$ commute. Thus, $\phi$ is a homomorphism, and $\phi((1,0)) = g$ and $\phi((0,1)) = h$. All that remains to be shown is that $\phi$ is unique. Suppose $\psi$ is another homomorphism $\Bbb Z\times\Bbb Z\to G$ such that $\psi((1,0)) = g$, $\psi((0,1)) = h$. Then $\psi((n,m)) = \psi(n(1,0) + m(0,1)) = \psi(n(1,0))\psi(m(0,1)) = g^n h^m,$ so $\psi = \phi$.