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 $1$.
[Math] Isomorphism between $U(15)$ and $\Bbb Z_4\oplus \Bbb Z_2$
abstract-algebragroup-isomorphismgroup-theory
Related Solutions
The general way to find all homomorphism $\mathbb Z_n\to G$ for an arbitray abelian group $G$ is the following: Suppose $\phi:\mathbb Z_n\to G$ is a group homomorphism, as you said, it is determined by the image of $1$, so the question really is which choices of $g\in G$ give a homomorphism $\mathbb Z_n\to G$ when picked as the image of $1$?
For homomorphisms $\psi:\mathbb Z\to G$, this is easy: Pick any $g\in G$, let $\psi(x)=xg$ and just check $$\psi(x+y)=(x+y)g=xg+yg=\psi(x)+\psi(y).$$
Now what could go wrong? The subtle thing that happens here is better exposed, when we write $G$ as a multiplicative group. The definition of $\psi$ becomes $\psi(x)=g^x$ and the calculation becomes $$\psi(x+y)=g^{x+y}=g^x g^y =\psi(x)\psi(y).$$ The important fact: $g^x$ is defined for $x\in\mathbb Z$, that's why $\mathbb Z\to G$ is easy.
Going back to $\phi:\mathbb Z_n\to G$ and sticking to multiplicative notation, it is tempting to just choose $\phi(1)=g$ and define $\phi(x)=g^x$. But here you run into trouble: When $x\in\mathbb Z_n$, what is $g^x$ supposed to mean? Remember that $x\in\mathbb Z_n$ really is a coset of $n\mathbb Z$ in $\mathbb Z$, consisting of all elements of the form $x+kn$ with $k\in\mathbb Z$ and that $\mathbb Z_n$ is really just a shorthand for $\mathbb Z/n\mathbb Z$, the set of all cosets equipped with addition.
What you are really doing is defining $\psi:\mathbb Z\to G$ by $\psi(x)=g^x$ (which is fine) and then look at the induced map $\phi:\mathbb Z_n\to G$ with $\phi(x+n\mathbb Z)=g^x$ for all $x\in\mathbb Z$. Now the definition depends on the representative $x$ of the coset $x+n\mathbb Z$. For the map to be even well defined, it has to be independent of the choice of the representative: If $x+n\mathbb Z=y+n\mathbb Z$, so $x-y=kn$ for some $k\in \mathbb Z$, we want $g^x=g^y$, so $1=g^{x-y}=g^{kn}$. Thus, for $\phi$ to be well defined, we need $g^{kn}=1$ for all $k\in\mathbb Z$: The order of $g$ needs to be a divisor of $n$. When this is the case, the same calculation as above reveals that $\phi$ is not only well defined, but a well defined group homomorphism.
Now that we know, that we can pick any $g\in G$ with order diving $n$, the example $\mathbb Z_4\to\mathbb Z_2\oplus\mathbb Z_2$ is a quick one: All elements of $\mathbb Z_2\oplus \mathbb Z_2$ are of order $1$ or $2$, both are divisors of $4$, so we can pick any of them.
Yes, you are correct!
Here is a relevant OEIS sequence entry. It agrees with your calculation.
Best Answer
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 $$ \phi(n,m)=2^n11^m. $$