Show that $U(8)$ is Isomorphic to $U(12)$. 
Question: Show that $U(8)$ is Isomorphic to $U(12)$

The groups are:
$U\left ( 8 \right )=\left \{ 1,3,5,7 \right \}$
$U\left ( 12 \right )=\left \{ 1,5,7,11 \right \}$
I think there is a bit of subtle point that I am not fully understanding about isomorphism which is hindering my progress. The solution mentions about the order of an element but I do not understand how that is pivotal to solving this.
Thanks in advance.
 A: The preferred approach would be to explicitly construct that isomorphism, but since the problem suggests basing the proof around the concept of "order of an element", let us do so then.
The crucial thing here is that both groups are of order $4$, and a simple theorem says that there exist only two classes of groups of order $4$: those isomorphic to $\Bbb Z_2 \times \Bbb Z_2$ (also known as "the Klein group") and those isomorphic to $\Bbb Z_4$. Notice that in $\Bbb Z_2 \times \Bbb Z_2$ all the elements are of order $2$, while in $\Bbb Z_4$ there are elements of order $4$.
It is now a matter of simple calculations to check that in your groups all the elements have order $2$, therefore they cannot be isomorphic to $\Bbb Z_4$, hence by the above considerations they must be isomorphic to $\Bbb Z_2 \times \Bbb Z_2$, hence isomorphic between themselves. (Notice that we don't have an explicit formula for this isomorphism - but neither are we required to find it.)
A: Let $F:U(8) \rightarrow U(12)$ be such that
$F(1)=1´;
F(3)=11´;
F(5)=5´;
F(7)=7´$. Note that $ F$ is bijective, to
show that $F$ is an isomorphism we only need to show that $F$ is indeed operation preserving.Observe
$F(1*n)=F(n) = F(n)*1'=F(n)*F(1)$ for all $n \in U(8)$;
$F(3*5)=F(7)=7'=11'*5'=F(3)*F(5);\\ F(5*7)=F(35)=F(3)=11´=5´*7´=F(5)*F(7); \\ F(3*7)=F(21)=F(5)=5´=11´*7´=F(3)*F(7). $
(Where $*$ reffers for respective group operations)
$\Rightarrow F$ is a homomorphism & F is bijective
$\Rightarrow F$ is an isomorphism
$\Rightarrow U (8) \cong U(12).$
