Consider the dihedral group $D_n = \{a^i b^j \mid 0 \leq i <6, \ 0 \leq j< 2 \}$ where $a^n = b^2 = 1$ , and $a b =b a ^{-1}$. For any divisor $k$ of $n$ show that
$$\langle a^k\rangle\triangleleft D_n \text{ and } D_n/\left \langle a^k \right \rangle\cong D_k $$
Answer:
From the definition $ D_n = \{a^i b^j \mid 0 \leq i <6, \ 0 \leq j< 2 \} $ , it is clear that $n=6$.
So we are dealing with $ D_6 \ $.
Now the divisors of $6$ are $ 1, \ 2, \ 3, \ 6 $.
So the subgroups are
$$ \langle a\rangle , \ \langle a^2\rangle , \ \langle a^3\rangle , \ \langle a^6\rangle $$
Further,
$$ \langle a\rangle =\{1,a,a^2,a^3,a^4,a^5 \}, $$
$$\langle a^2\rangle =\{1, a^2,a^4\}, $$
$$ \langle a^3\rangle =\{1,a^3 \}, \\ \langle a^6\rangle =\{1\} $$
Now, $|D_6|=12$
Then,
$$ \frac{|D_6|}{|\langle a\rangle|}=\frac{12} 6 = 2, $$ So the index $ [D_6 : \langle a\rangle ]=2.$
Hence $\langle a\rangle $ is a normal subgroup of $D_6$.
But I am unable to prove $D_n /\langle a^k\rangle \cong D_k \ $ . Is there any help ?
$\require{begingroup}\begingroup\newcommand{\wt}{\widetilde}$Let us say
$$D_r = \langle a_r, b_r \mid a_r^r = 1, b_r^2 = 1, a_rb_r = b_ra_r^{-1} \rangle. $$
The subscript $r$ lets us distinguish between elements of $D_n$ and $D_k$.
Let $\wt x$ denote the image of $x \in D_n$ modulo $\langle a^k \rangle$. Then
This allows us to say that the map $D_k \to D_n / \langle a^k \rangle$ defined by
$$ a_k \mapsto \wt{a_n} \text{ and } b_k \mapsto \wt{b_2} $$
is a surjective homomorphism (check this). But $|D_k| = 2k = |D_n / \langle a^k \rangle|$, so this has to be an isomorphism.
Your proof that $\langle a^k \rangle$ is normal isn't a proof of the intended statement. If there was a $6$ in your assignment, that's a typo and should be an $n$. So you can't assume $n = 6$. You are also assuming that $k = 1$, which isn't general enough.
$\endgroup$