Let's look at the structure of $(\Bbb Z_7)^{\times}$ in some more detail.
$[1]$ isn't very interesting, it's clearly the (multiplicative) identity, though.
So now let's look at $[2]$, specifically, its powers.
$[2]^2 = [2]\cdot [2] = [4]$. This is...boring.
$[2]^3 = [2]\cdot[2]\cdot[2] = [8] = [1]$ (because $8 = 1 + 7$, so $8 \equiv 1$ (mod $7$)).
This tells us $[2]$ has order $3$.
OK, so now let's look at $[3]$.
$[3]^2 = [9] = [2]$. Thus $[3]^6 = ([3]^2)^3 = [2]^3 = [1]$. So the order of $[3]$ divides $6$ (this is, of course, self-evident by Lagrange).
As we saw above $[3]^2 = [2] \neq [1]$, so the order of $[3]$ is NOT $2$.
$[3]^3 = [27] = [6] \neq [1]$, so the order of $[3]$ is NOT $3$.
This means the order of $[3]$ must be $6$, so it generates the entire group:
$[3]^1 = [3]$
$[3]^2 = [2]$
$[3]^3 = [6]$
$[3]^4 = [4]$ ($81 = 4 + 7\ast 11$)
$[3]^5 = [5]$ ($243 = 5 + 7\ast 34$)
$[3]^6 = [1]$
Therefore, an isomorphism between $(\Bbb Z_6,+)$ and $((\Bbb Z_7)^{\times},\ast)$ is:
$[k]_6 \mapsto [3^k]_7$ (where the subscripts/brackets mean "equivalence class modulo").
Note $[3]^{-1} = [5]$, so we could have used $[5]$ as a generator instead.
Best Answer
In the group $U$ there are more than two elements whose order is finite namely, $i,-i,1,-1$. But in other two groups, it's not true. As in $(\mathbb{R}^*,\times)$ only $\pm1$ are of finite order and in $(\mathbb{R},+)$ there is no element of finite order.