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

  • 1
    $\begingroup$ What does "x21" mean? $\endgroup$ – user35603 May 11 '14 at 13:42
  • $\begingroup$ x(subscript) 21 is the group operation $\endgroup$ – Bob May 11 '14 at 13:55
  • 2
    $\begingroup$ And what is the definition of this operation?... $\endgroup$ – Ludolila May 11 '14 at 14:09
  • $\begingroup$ Multiplication modulo 21 I would guess. If so, what is the multiplicative identity OP? $\endgroup$ – ah11950 May 11 '14 at 14:25
  • $\begingroup$ @ah11950: it seems to $15$ ? $\endgroup$ – 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.


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