Prove or disprove that $U(8) \cong Z_4$ 
Prove or disprove that $U(8) \cong Z_4$

I understand that $U(8)=\{1,3,5,7\}$ a group under the operation multiplication mod8 and $Z_4 = \{0,1,2,3\}$ a group under the operation addition mod4.
Given the function $\Phi: U(8) \longrightarrow Z_4$, I chose the following bijection:
$$ 1 \longrightarrow 0 $$
$$ 3 \longrightarrow 1$$
$$ 5 \longrightarrow 2$$
$$ 7 \longrightarrow 3$$
I want now to show that the group operations are preserved. Given $a,b \in U(8)$, Let's check that $$\Phi(ab)mod8=[\Phi(a) +\Phi(b)] mod4$$
I did check with several values and it shows that the group operations are preserved.
Question 1: Is this correct to stipulate that the result of the Left-Hand-Side to be evaluated in mod8, and the result on the Right-Hand-Side in mod4?
Question 2: To prove that the group operations are preserved, do I have to show the 6 possible combinations of $a$ and $b$ $\in U(8)$?
Question 3: Is there another method to show that there is an homomorphism from $U(8)$ to $Z_4$ that is more appropriate or cleaner?
Question 4: The mapping could be different under another function $\Psi$ for example. If I show that group operations are preserved with $\Phi$, will the group operation under $\Psi$ be necessarily preserved?
I am a bit stock, I think because of those questions. Much appreciated.
 A: you may notice that your group $U(8)$ has three elements of order two, together with the identity. $Z_4$ on the other hand has two elements of order $4$. So the groups cannot be isormorphic. 
A: There are lots of pieces here. Let's go through a couple of them very slowly and clearly.
First, suppose we have $a$ and $b$ in $U(8)$. Then as $U(8)$ is a group under multiplication mod 8, by $ab$ we mean the (honest, good old integer) product of $a$ and $b$ modulo $8$.
On the other side, $Z_4$ is a group under addition mod $4$.
You have proposed a bijection $\Phi: U(8) \longrightarrow Z_4$.
You should ask: is this bijection a homomorphism? One property of a homomorphism is that $\Phi(a^2) = \Phi(a)^2$, where $a^2 = a\cdot a$ (as this is the group product in $U(8)$) and $\Phi(a)^2 = \Phi(a) + \Phi(a)$ (as this is the group product in $Z_4$). Is it true that $\Phi(a^2) = \Phi(a) + \Phi(a)$ for your proposed bijection?
No. In fact, this isn't true for any of the elements other than the identity element $1$.
A: *

*You need $\phi(ab \mod 8) = [\phi(a) + \phi(b)] \mod 4.$

*Ideally you would show that the operation is preserved for an arbitrary pair of elements. Since $\phi$ is claimed to be an isomorphism of finite groups it is possible to check each element individually, but generally there's a better way.

*Since you have defined the claimed isomorphism elementwise, there isn't. If you find an isomorphism which can be expressed in terms of arbitrary elements, sure. 

*I don't see why you think they would. If $\psi$ is different from $\phi$ it need not even be an isomorphism. 
