Check if $U(20)$ is isomorphic to $U(24)$ We have: 
$U(20) = {\{1,3,7,9,11,13,17,19\}}$ 
$U(24) = {\{1,5,7,11,13,17,19,23\}}$ 
We see that $|U(20)| = |U(24)|$, so it's not as simple as just checking if the orders differ to prove they're not isomorphic. I know that they are not, but how would I do that without computing the orders of every element in both groups to see if there's a one to one mapping between elements of the same order between sets?
 A: Hint: $U(20)$ contains an element of order $4$ but $U(24)$ does not.
A manifestation of this fact is that the map $x \mapsto x^2$ is the trivial map on $U(24)$ but not on $U(20)$, and this is easy to check.
A: By the Chinese Remainder theorem,
$\;\mathbf Z_{20}^\times\simeq\mathbf Z_{4}^\times \times\mathbf Z_{5}^\times$, and the latter groups are cyclic, therefore $\;\mathbf Z_{20}^\times$ is isomorphic to the (additive) group  $\mathbf Z_{2}\times\mathbf Z_{4}$.
On the other hand, $\;\mathbf Z_{24}^\times\simeq\mathbf Z_{3}^\times \times\mathbf Z_{8}^\times$. Now, $\mathbf Z_{3}^\times $ is cyclic, isomorphic to $\mathbf Z_{2}$, but $ Z_{8}^\times$ is not cyclic: it is the internal direct product of the subgroup of order $2$ generated by the congruence class of $-1$ and the subgroup, also of order $2$, generated by the congruence class of $5$, hence it is isomorphic to $\mathbf Z_{2}\times \mathbf Z_{2}$, so that, ultimately
$$\mathbf Z_{24}^\times\simeq \mathbf Z_{2}\times \mathbf Z_{2}\times\mathbf Z_{2}.$$
A: So $\mathbb Z$ and $\mathbb Z_n = \{0, 1, ... , n-1\}$ are examples of rings.  Rings have addition, multiplication, zero, and one.  If $R$ is a ring, the units $U(R)$ is by definition the set of all elements $x \in R$ such that there exists a $y \in R$ satisfying $xy = 1$.  Then $U(R)$ is a group with respect to multiplication.
By definition $U(n)$ is $U(\mathbb Z_n)$, the units in $\mathbb Z_n$.  For example, $\mathbb Z_4 = \{0, 1, 2, 3\}$, and $U(4) = \{1,3\}$.  
If $R_1$ and $R_2$ are rings, then the Cartesian product $R_1 \times R_2$ is also a ring.  Also, $U(R_1 \times R_2) = U(R_1) \times U(R_2)$.  
You can put all this together to solve your problem. So $\mathbb Z_{20}$ and $\mathbb Z_4 \times \mathbb Z_5$ are isomorphic as rings by the Chinese remainder theorem.  Then by what I said above, $U(20)$ and $U(4) \times U(5)$ are isomorphic as groups.
Similarly, $\mathbb Z_{24} \cong \mathbb Z_3 \times \mathbb Z_8$, so $U(24) \cong U(3) \times U(8)$.  
So if $U(20)$ and $U(24)$ were isomorphic, then $U(3) \times U(8)$ and $U(4) \times U(5)$ would be isomorphic.  Think about why this is impossible.
