Understanding cyclic groups with Cayley tables

Question. Let $G = \{a,b,c,d,f\}$. Given that $(G, \cdot)$ is a cyclic group with $G=\langle d \rangle$ and Cayley table:
\begin{array}{c|cc} \cdot & a & b & c & d & f\\ \hline a& c & a & f & b & d \\ b& a & b & c & d &f \\ c& f& c& d& a& b \\ d& b & d& a& f & c \\ f& d& f& b& c & a \end{array}

I need to complete this table. I know that the generator will be all the powers of d such that $d^1 ... d^4 ∈ G$, where $n ∈ Z$. My understanding of that statement is each row and column of d will contain each element of G only once.

I know cyclic groups are abelian, but I only used the commutative property thus far to fill in 2 cells.

What else do I need to know in order to fill in the table?
Thank you.

• I'm the \langle \rangle fairy, here to let you know that $\langle, \rangle$ plays nicer with TeX than <, > does :) – Patrick Stevens Feb 25 '18 at 22:37
• @PatrickStevens thanks \langle \rangle fairy :)! – Charles Grealy Feb 25 '18 at 23:21

Note that $G$ has order $5$, hence all elements of $G$, other than $1$, have order $5$, and consequently, $x^5=1$, for all $x\in G$.

You have $d^2=f$, and $f^2=a$, hence $d^4=a$.

Then from $ad=b$, we get $a^5=b$, but $a^5=1$, hence $b=1$.

Since $b=1$, we have $d\ne 1$, so the order of $d$ must be $5$.

Hence, the elements $1,d,d^2,d^3,d^4$ are distinct, and comprise all the elements of $G$.

The only unidentified one is $d^3$, which must be equal to $c$.

Note: The table shows $df=c$, so we could have found $d^3=c$ from that information, but as the above argument shows, that information is actually superfluous. In other words, if the two entries of $c$ in the table of products were erased, we could still have deduced them.

To fill in the rest of the table, products can be evaluated as follows . . .

1. Express the factors as powers of $d$.$\\[4pt]$
2. Multiply the powers using the laws of exponents.$\\[4pt]$
3. Reduce the exponent of the result, mod $5$.$\\[4pt]$
4. Express the result as the unique element of the set $\{a,b,c,d,f\}$ whose exponent, as a power of $d$, matches the reduced exponent obtained for the product.
• Glorious! I was able to figure the rest out (I hope). I do have a question that may be obvious: does d^1 = d that is in G? So for b⋅d = 1⋅d = d – Charles Grealy Feb 25 '18 at 23:20
• Yes, for any group $G$, and for all $x\in G$, we have $x^1=x$, by definition. – quasi Feb 25 '18 at 23:27