abstract-algebrafinite-groupsgroup-theorymodular arithmetic
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?
Generally these tables are interpreted as taking an $x$ from the left column and a $y$ from the top row and putting its product $xy$ in the $(x,y)$ position.
You have already been told that $d$ goes in the $(a,b)$ position, and that $e$ goes in the $(c,a)$ position, and that $b$ goes in the $(d,c)$ position. Since groups of prime order are abelian, you can also conclude what $ba,cd$ and $ac$ are.
Since $a,b,c,d,e$ are likely assumed to be distinct, you can also tell from these that the only candidate for the identity is $e$, and so that allows you to fill in the last column and the last row rapidly. At this point you also learn that $a$ and $c$ are inverses of each other.
Using that relationship, you can deduce from $ab=d$ that $b=cab=cd$, so another entry appears in the $(c,d)$ position. As you get further along, you should be able to deduce each position.
Don't forget also that you have another tool at your disposal, namely that all the elements satisfy $x^5=e$. Another thing is that $a,c$ are paired up as inverses, and $e$ is its own inverse... what can you conclude about $b$ and $d$? Also, show that $a^2\in\{b,d\}$: if you try both of them out, you should see immediately that only one is consistent with the relations.
Please update us with your progress.
I think all you need has been pointed here. You don't need to write down the associated Cayley table for the presented set and its operation, but for this one it leads you to get the answer graphically:
We can see that the operation is a binary one. Can you find the identity element? What about the inverses ones?
Best Answer
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.