Group operation $G,\times_{21}$ with the set set $G=\{3,6,9,12,15,18\}$

Show that the set $G=\{3,6,9,12,15,18\}$ is a group under the operation $\times_{21}$. You should state the inverse of each element in $(G,\times_{21})$.

I'm sure $G_1$-closure $G_2$-Identity and $G_3$-Inverses hold although correct me if I'm wrong, but could someone please show me if $G_4$-Associativity holds. Thanks

• What does "x21" mean? – user35603 May 11 '14 at 13:42
• x(subscript) 21 is the group operation – Bob May 11 '14 at 13:55
• And what is the definition of this operation?... – Ludolila May 11 '14 at 14:09
• Multiplication modulo 21 I would guess. If so, what is the multiplicative identity OP? – ah11950 May 11 '14 at 14:25
• @ah11950: it seems to $15$ ? – mesel May 11 '14 at 14:36

If by $\times_{21}$ you mean multiplication modulo $21$, then you can use the fact that this operation is associative (even in the more general setting of $\mathbb Z_n$). See for example here: Multiplication group modulo n is well defined,associative.
Computing powers of 3 (mod 21) you get ${3, 9, 6, 18, 12, 15}$ and then back to 3 again. That makes the set of numbers a cyclic group of order 6 generated by 3 with $3^6 = 15$ the identity.