Answer: Let $D_8=\langle a,b\rangle=\{e,a,a^2,a^3,b,ba,ba^2,ba^3\}$, where $a^4 = b^2 = e$ and $ab = ba^{-1}$ as usual.
Then
$$\{e\}\lhd\langle a^2\rangle\lhd\langle a\rangle\lhd D_8$$ or
$$\{e\}\lhd\{e,a^2\}\lhd\{e,a,a^2,a^3\}\lhd\{e,a,a^2,a^3,b,ba,ba^2,ba^3\}$$ is a subnormal series and all three factors have order two.
Since any group of order two is simple, this is a composition series
Question: Can someone explain how the three factors have order two.
Factors: quotient groups $G_1/G_0,G_2/G_1,...,G_n/G_{n-1}$.
Quotient group: $G/H$ for $H\leq G$ is the set of distinct left cosets that for an element of $x\in G$ and $x\not\in H$, the left coset is defined as $xH=\{xh\mid h\in H\}$
For $\{e,a^2\}/\{e\}$, we have $x=a^2$ and this is the set: $\{ea^2\}=\{a^2\}$ (order 1)
For $\{e,a,a^2,a^3\}/\{e,a^2\}$, we have $x=a,a^3$ giving $\{ae,aa^2,a^3e,a^3a^2\}=\{a,a^3\}$ (order 2)
For $\{e,a,a^2,a^3,b,ba,ba^2,ba^3\}/\{e,a,a^2,a^3\}$ I get $\{b,ba,ba^2,ba^3\}$ (order 4)
Where am I going wrong, I should have factors all of order $2$ but have factors of order $1,2$ and $4$
The left $\{e,a^2\}$ cosets of $\{e,a,a^2,a^3\}$ are $$e\{e,a^2\},a\{e,a^2\}.$$ In particular there two of them. You can think of $\bar{a},\bar{e}$ as the elements of the quotient group, but you can also take other representatives of the cosets.