Prove that this set and operation comprise a group. Show that the set $\{5, 15, 25, 35\}$ is a group under multiplication modulo $40$. Is there a relation between this and $U(8)$?
I am having a really difficult time beginning this proof. All this is abstract to me. 
 A: Hint: Since there are only four elements (and the group is abelian because multiplication is commutative), it's easy to write down the multiplication table and check the axioms directly.
For example, one finds that $25 \cdot g = g$ for all $g$ in the set, so $25$ is the identity.
For the associativity axiom, one need not use the multiplication table, as we already know the group operation (multiplication modulo a given number) is associative.
There is a relationship between this and the ring $U(8)$. Namely, we can consider the group $U(8)^* = \{1, 3, 5, 7\}$ of invertible elements in $U(8)$, and this turns out to be isomorphic to the given group (both are isomorphic to the Klein $4$-group, $U(2) \times U(2)$).
A: Well, it is abstract algebra... Jokes aside, do you know the axioms of a group? You need to check, for each pair of elements in the set, that the product mod $40$ is still in the set, find the identity (should surface when you check for closure), and an inverse for each element. For a set like this one, you can do each calculation by hand (there are after all just three possibilities for each inverse). The associativity law follows from the associativity of integers.
Once you have that it is a group, you can set up the multiplication table, and compare it to the multiplication table of $U_8$. See if you can spot any similarities.
Comment if you need me to fill in more details!
