Multiplication modulo proving a set is a group conform conditions Given is the following explanation.
A group is a set, together with a binary operation $\odot$, such that the following conditions hold:


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*For all a, b $\in$ S it holds that a $\odot$ b $\in$ S


Now, i have to show that the Set = {1, 2 ... 16, 17} together with multiplication modulo 18 is not a group.
 A: Multiplication modulo $14$ means the value of $a\cdot b \pmod {14}$. So for $a = 1, b = 2$, $a\odot b = 1 \cdot 2 \pmod {14} = 2.$ This is an associative operation on the integers, because multiplication is associative.
HINT for the first: The identity element here is $1$. Show that $2$ has no multiplicative inverse $a'$ (no element in the group such that  $2\cdot a' = a'\cdot 2 = 1$. Indeed, you can simply show that your set is not closed under the binary operation $\odot$: Calculate for example, $2 \odot 7 =  2\cdot 7 \pmod{14}$.

HINT for the second: Note that $13$ is prime, and any group $\mathbb Z^*_{p}$ where p is prime has $p - 1$ elements, and is equal to $\{1, 2, \cdots, p-1\}$. 
$1$ again is your identity element: this is easy enough to confirm. You can also verify that the inverse $a'$ of every element $a\in \mathbb Z^*{13} \in \mathbb Z^*_{13}$. You can also confirm that for any two elements $a, b \in \mathbb Z^*_{13}$, $a \odot b = a\cdot b \pmod {13} \in \mathbb Z^*_{13}.$
A: Hint for your first question: compute $2\odot7$.
