Prove that $U_8$ is isomorphic to a group of matrices 
Prove that $U_8$ is isomorphic to the group of matrices 
  $$G=\left\{\pmatrix{1 & 0 \\ 0 &  1}, \pmatrix{1 & 0 \\ 0 & -1},\pmatrix{-1 & 0 \\ 0 & 1},\pmatrix{-1 & 0 \\ 0 & -1}\right\}.$$

I was thinking to create a bijection $\phi$ between the set $U_8$ and $G$ then prove that the operation is preserved using $\phi(f \circ g)= \phi(f) \cdot \phi(g)$. But I would have to check for all combination of objects and then for all possible bijections. It does not look appropriate to proceed as such.
I noticed also that the order of the non-identity elements in each finite group is 2. But, I do not know what to do out of this.
I draw the Cayley tables for each group. They show that both groups are abelian. Both also respect the same structure such as that, for example, the product of the second element and the 3rd element equals to the 4th element in both $U(8)$ and $G$. Again I am a bit puzzled; it does seem that something is missing.
Any input is much appreciated
 A: I believe this was already answered here
Show U(8) is isomorphic to the group of matrices...
That being said I found the answer there to not have the most depth, so here's my (hopefully correct) take:
To prove the groups are isomorphic, we must prove a $\phi$ homomorphic and bijective. Let's define the group of matrices as $G=\left\{\left(\matrix{1&0\\0&1}\right),\left(\matrix{1&0\\0&-1}\right),\left(\matrix{-1&0\\0&1}\right),\left(\matrix{-1&0\\0&-1}\right)\right\}$. To be clear, $U(8)$ is defined as $U(8)=\left\{1,3,5,7\right\}$ where $1=3^2=5^2=7^2$.
Homomorphic
Let's define $\phi:U(8)\to G$ as
$\phi(1)=\left(\matrix{1&0\\0&1}\right)$, $\phi(3)=\left(\matrix{1&0\\0&-1}\right)$, $\phi(5)=\left(\matrix{-1&0\\0&1}\right)$, $\phi(7)=\left(\matrix{-1&0\\0&-1}\right)$
I chose these translations as
$5\times7=35\mod(8)=3$, and $\left(\matrix{-1&0\\0&1}\right)*\left(\matrix{-1&0\\0&-1}\right)=\left(\matrix{1&0\\0&-1}\right)$
$3\times7=21\mod(8)=5$, and $\left(\matrix{1&0\\0&-1}\right)*\left(\matrix{-1&0\\0&-1}\right)=\left(\matrix{-1&0\\0&1}\right)$
$3\times5=15\mod(8)=7$, and $\left(\matrix{1&0\\0&-1}\right)*\left(\matrix{-1&0\\0&1}\right)=\left(\matrix{-1&0\\0&-1}\right)$
Where $u_1,u_2\in U(8)$, and $u_1\not=u_2$ (the case where $u_1=u_2$ is trivial as it would result in the identity which is shared), thus $\phi(u_1\times u_2)=\phi(u_3)$ such that $u_3\in U(8)$ where $u_3\not=u_1$, $u_3\not=u_2$, and $u_3\not=e$ as $u_1\not=u_2$. From $\phi(u_3)=g_3$ such that $g_1,g_2,g_3\in G$ where $g_3=g_1*g_2$ as $g_3\not=e$, thus $\phi(u_3)=g_3=g_1*g_2$. As $g_1,g_2$ are unique and not equal to $g_3$ or the identity, then $g_1=\phi(u_1)$ and $g_2=\phi(u_2)$ for some unique $u_1\not= u_3$ and $u_2\not= u_3$ (where $u_1,u_2\not=e$). Thus we have found that  $\phi(u_1\times u_2)=\phi(u_3)=g_3=g_1*g_2=\phi(u_1)*\phi(u_2)$ proving that $\phi$ homomorphic.
Bijective
One-to-one
This is immediately clear given our definition of $\phi$.
Onto
This is also immediately clear given our definition of $\phi$.
Given $\phi(u_1)=\phi(u_2)$ where $u_1,u_2\in U(8)$, and $\phi(u_1)=g_1, \phi(u_2)=g_2$ where $g_1,g_2\in G$ then $\phi(u_1)=\phi(u_2)\iff g_1=g_2$.
Bijectivity Conclusion
As $\phi$ is one-to-one and onto, it is bijective.
Isomorphic Conclusion
As we have found $\phi$ to be homomorphic and bijective for $\phi:U(8)\to G$ then $U(8)$ is isomorphic to the group of matrices $$\left(\matrix{1&0\\0&1}\right),\left(\matrix{1&0\\0&-1}\right),\left(\matrix{-1&0\\0&1}\right),\left(\matrix{-1&0\\0&-1}\right)$$
A: The way I would do it is to check that they're each noncyclic groups of order $4$.
$U_8$ has order $\varphi (8)=4$, and no element of order $4$. Thus it's $V_4$.
Similarly check that your matrix group has order $4$, and is not cyclic.
As you noted,  proving directly that you have an isomorphism appears slightly cumbersome.
A: As there are only two groups of order $4$ (up to isomorphisms), if you know that $U_8\cong C_2\times C_2$ (otherwise confirm that it hasn't got any element of order $4$), then you are left to check whether $G$ has an element of order $4$, or not. You will find that it hasn't got any, so necessarily $G\cong U_8(\cong C_2\times C_2)$.
