Isomorphism between $U(15)$ and $\Bbb Z_4\oplus \Bbb Z_2$ I am having trouble finding an isomorphism between $U(15)$ and $\mathbb Z_4\oplus \mathbb Z_2$. What we learned in class to create an isomorphism we typically find the generators and go from there. However, I can't find a generator for wither group. I am supposed to list where each element of $U(15)$ is mapped and also check three cases such that $\varphi(ab) = \varphi(a)\varphi(b)$. I know the elements in $$U(15)=\{1,2,4,7,8,11,13,14\}$$ and in $$\mathbb Z_4\oplus \mathbb Z_2=\{(0,0),(0,1),(1,0),(1,1),(2,0),(2,1),(3,0),(3,1)\}.$$ I found the order of $(0,0)=1$,$(0,1)=2$, $(1,0)=4$, $(1,1)=4$,$(2,0)=2$,$(2,1)=2$,$(3,0)=4$,$(3,1)=4$. In $U(15)$ $4,11,14$ have order $2$ and $2,7,8,13$ have order $4$ and $1$ has order $1$.

You have that $\mathbb Z_4\oplus\mathbb Z_2$ is generated by two commuting elements, one of order $4$ and one of order $2$ (namely, $(1,0)$ and $(0,1)$). 
On $U(15)$, we want generators $a$ of order $4$ and $b$ of order $2$; if we take $a=2$, the table below shows that $b\ne4$; taking $b=11$,
$$
\begin{matrix}
a=2\\
a^2=4\\
a^3=8\\
b=11\\
ab=7\\
a^2b=14\\
a^3b=13
\end{matrix}
$$
So $a,b$ generate $U(15)$ and we can define $$
\phi(n,m)=2^n11^m.
$$
