Multiplication modulo proving a set is a group Show that set $\{1,2,3\}$ under multiplication modulo $4$ is not a group but that $\{1,2,3,4\}$ under multiplication modulo $5$ is a group.
 A: I will show that $\{1,2,3\}$ is not a group under multiplication mod $4$ and you can verify that $\{1,2,3,4\}$ is a group under multiplication mod $5$.
As was given as a hint, all we need to do is make a multiplication table and investigate it.
$$
        \begin{array}{c | c c}
            \times & 1 & 2 & 3 \\
    \hline
            1 & 1 & 2 & 3 \\
            2 & 2 & 0 & 2 \\
            3 & 3 & 2 & 1
        \end{array}
$$
Do you see why this is not a group?  What do you know about the Cayley tables of groups?
Hint: How many times can an element show up in each row/column?
After edit: Alternatively, the multiplication on the group is a binary operation which means it must be closed, so having $0$ show up is not possible.
A: Try and provide some work so we can see where you are stuck or any misconceptions you might have that we might help correct.
First, realize that in order for a set and and an operation $*$ to qualify as a group $G$, the set and the operation must satisfy the four group axioms:
1 - Closure: For $a,b \in G$, $a*b \in G$.
2 - Associativity: For $a,b,c \in G$, $(a*b)*c=a*(b*c)$.
3 - Identity element $e$: There exists an element $e$ in $G$ such that for each $a \in G$, $e*a=a*e=a$.
4 - Inverse element: For each $a$ in $G$, there exists an element $b$ in $G$ such that $a*b=b*a=e$.
Also, multiplication modulo 4 is essentially first multiplying the numbers and then taking the result modulo 4. Example: 2 multiplied by 5 modulo 4 is congruent to 2 since $2 \cdot 5 = 10 \equiv 2$ mod $4$.
Now you understand the requirements of the problem, and you must verify that the set $\{1,2,3\}$ under multiplication modulo 4 does not satisfy the four group axioms. This can be done as follows. By 3, the identity element must be $1$. Taking a look at number 4 about the inverse element, it is clear that there doesn't exist an inverse element for $2$ or $3$: 2 times 3 is congruent to $2$ mod $4$, which is different from the identity (2 and 3 are the only elements of the set different from the identity, and so they must be inverses of each other, which we have just shown that they are not). This proves the first claim.
Now prove that 1-4 holds for the second set under multiplication modulo 5.
