Generally speaking: suppose there was a generator and derive a contradiction.
(Alternative methods include, for example, showing the group is isomorphic to another group which you already know to be non-cyclic.)
For $G_2$, suppose $a$ is the generator. Will you ever be able to get the rational number $3a/2$? (That is, $a^1 = a$ and $a^2 = a + a = 2a$; but what about the point halfway between $a$ and $2a$?)
For $G_3$, suppose without loss of generality that $a > 1$ is the generator. (If $b < 1$ is a generator, then so is $b' = 1/b > 1$.) Then we can generate $a$ and $a^2$, but what about the point halfway between them? That is, what about $(a + a^2)/2$?
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 . . .
- Express the factors as powers of $d$.$\\[4pt]$
- Multiply the powers using the laws of exponents.$\\[4pt]$
- Reduce the exponent of the result, mod $5$.$\\[4pt]$
- 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.
Best Answer
Is your question "Given a group $(G,*)$ and an element $a \in G$, how do we define the subgroup generated by $a$ (denoted $<a>$)" ?
If so, since $G$ is a group, every element has an inverse, hence $a^{-1}$ is just the inverse of $a$ in the group $(G,*)$ and $a^{-n}= a^{-1}* \dots *a^{-1}$ is just the multiplication of $a^{-1}$, $n$ times.
By convention $a^0:=e$ is the neutral element of $G$.
$<a>= \{ a^n, \ n \in \Bbb Z \}$ is a well defined subgroup of $G$ (and it's easy to see that it's the smallest subgroup containing $a$).
We say that $a$ generates $G$ if $G= <a>$.