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Does the set $H=\{1,4,7,13\}$ with modulo $15$ multiplication, $\otimes_{15}$, create a group?

$$\begin{array}{|r|c|c|c|} \hline \otimes_{15} & 1 & 4 & 7 & 13\\ \hline 1 & 1 & 4 & 7 & 13\\ \hline 4 & 4 & 1 & 13 & 7\\ \hline 7 & 7 & 13 & 4 & 1\\ \hline 13 & 13 & 7 & 1 & 4\\ \hline \end{array}$$

I learned that I have to make a table. What can I read from it?

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You just need to verify the axioms of the definition of a group for $\mathscr{H}=(H, \otimes_{15}).$

The set $H$ is closed under $\otimes_{15}$ by inspection of the multiplication table. (It satisfies the Latin square property.)

The identity is $1$.

The inverse of $4$ is itself. The inverse of $7$ is $13$ and vice versa.

Associativity of $\otimes_{15}$ is inherited from that of ordinary multiplication.

Hence $\mathscr{H}$ is a group.

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  • $\begingroup$ is there a different approach to this task without using the table? $\endgroup$ – vmahth1 Oct 9 at 9:08
  • $\begingroup$ @vmahth1: see my answer $\endgroup$ – J. W. Tanner Oct 9 at 23:00
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Yes, it is the subgroup of the group of units modulo $15$ that is generated by $7$ (or $13$).

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