I have the non-abelian group, the dicyclic group ${\rm Dic}_n$. I want to find a composition series such that the factor groups are abelian.
This is what I know:
The Dicyclic group is non-abelian. The composition factors must be simple and are unique up to order and isomorphism.
I also know that ${\rm Dic}_n$ can be shown as the set $\lbrace 1, a, a^2, \ldots, a^{2n-1}, x, ax, a^2x, \ldots, a^{2n-1}x\rbrace$.
For a group $G$ to have composition series to have abelian composition factors $1 = a \lhd b = G$, I have to show $a$ is a normal subgroup of $b$ and that $b/a$ is simple and $b/a$ is abelian. Since $x^2 = -1$ and $a^n = -1$ , I believe that the composition factors have something to do with $a^n$ and $x$ but I don't know exactly how they are simple subgroups.
Any help would be appreciated.
It is unconventional to refer to groups and subgroups by lowercase letters, and it is overloading the letter $a$ to use it to refer both to an element and a subgroup.
It is instructive to first examine composition series for cyclic groups. Let's look at $G=\mathbb{Z}/n\mathbb{Z}$, the integers mod $n$ under addition. This is cyclic of order $n$ generated by $1$. You should prove there is a unique subgroup of order $d$ for every divisor $d\mid n$, cyclically generated by $n/d$ (or technically $n/d+n\mathbb{Z}$). Thus, you can find a composition series $G_0\triangleleft G_1\triangleleft G_2\triangleleft\cdots\triangleleft G_m$ where $m$ is the number of prime factors of $n$ counted with multiplicity, by picking $|G_0|,|G_1|,|G_2|,\cdots,|G_m|$ be a sequence of divisors of $n$, each containing exactly one more prime divisor than the next, necessarily starting with $|G_0|=1$ and ending with $|G_m|=n$. This ensures the quotients $G_{k+1}/G_k$ have prime order, hence are simple subquotients. Indeed, we can order the prime divisors of $n$ (again, counted with multiplicitiy), and then turn that into the sequence of $|G_k|$s.
Let's do an example, $G=\mathbb{Z}/12\mathbb{Z}$. Its order factors as $12=2^2\cdot3$, and we get three possible ways to order its prime factors: (i) $2,2,3$; (ii) $2,3,2$; (iii) $3,2,2$. This gives three sequences for $|G_0|,|G_1|,|G_2|$ - namely, (i) $1,2,4,12$; (ii) $1,2,6,12$; (iii) $1,3,6,12$. The corresponding composition series for $\mathbb{Z}/12\mathbb{Z}$ are
$$ \langle 12\rangle \xrightarrow{\mathbb{Z}_2} \langle 6\rangle \xrightarrow{\mathbb{Z}_2} \langle 3\rangle \xrightarrow{\mathbb{Z}_3} \langle 1\rangle $$
$$ \langle 12\rangle \xrightarrow{\mathbb{Z}_2} \langle 6\rangle \xrightarrow{\mathbb{Z}_3} \langle 2\rangle \xrightarrow{\mathbb{Z}_2} \langle 1\rangle $$
$$ \langle 12\rangle \xrightarrow{\mathbb{Z}_3} \langle 4\rangle \xrightarrow{\mathbb{Z}_2} \langle 2\rangle \xrightarrow{\mathbb{Z}_3} \langle 1\rangle $$
It makes more sense to call $\langle 12\rangle$ by $\langle0\rangle$, but by writing $12$ above you can see a pattern in the least positive representatives for the subgroups' cyclic generators: you get them by dividing by primes.
Now for $\mathrm{Dic}_n=\langle a,x \mid a^{2n}, x^2=a^n,xax^{-1}=a^{-1}\rangle$. Since $N=\langle a\rangle$ has index $2$, it is normal. It doesn't contain $x$, so we can say $G/N=\langle xN\rangle$ is cyclic of order $2$ generated by $x$'s coset. Then you can construct a composition series $N_0\triangleleft N_1\triangleleft\cdots \triangleleft N_m$ for $N$, and we get a composition series
$$ N_0\triangleleft N_1\triangleleft N_2\triangleleft \cdots\triangleleft N_m\triangleleft G. $$
The composition factors are $\mathbb{Z}/p\mathbb{Z}$ for prime factors $p\mid n$, counted with multiplicity, with a (possibly extra) factor of $\mathbb{Z}/2\mathbb{Z}$ for $G/N$.