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

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?

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.

• is there a different approach to this task without using the table? – vmahth1 Oct 9 at 9:08
• @vmahth1: see my answer – J. W. Tanner Oct 9 at 23:00

Yes, it is the subgroup of the group of units modulo $$15$$ that is generated by $$7$$ (or $$13$$).