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Let G= $D_{20}$ note its the one that has 40 elements. Let $ H= <r^4> $

I)Prove that H is a normal subgroup.

II) Prove that $ G/H $ is not abelian

Now i am a lazy person and though im sure i could use $ghg^{-1} \in H \space \space\forall g \in G $ i would like to do is the following way if possible.

Define $\phi:G \to S $ where $ \phi (g) \to g^{mod 4} $ where S is basically just $D_4 $ clearly the map is surjective since

$\phi (r) \to r$

$\phi (r^2) \to r^2$

$\phi (r^3) \to r^3$

$\phi (r^4) \to e$

$\phi (rj) \to rj$

$\phi (r^2j) \to r^2j$

$\phi (r^3j) \to r^3j$

$\phi (r^4j) \to j$

I would like to say that $\phi $ is a homomorphism and $ker \phi = H $ so H is a normal subgroup. ( not sure how to prove it is an homomorphism )

Then i would like to apply the First isomorphism theorem to S or $G/H$ to show that it is isomorphic to $D_4$ and $D_4$ is clearly not abelian.

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There is one basic equation that helps a lot to understand dihedral group $D_n$: $$rs=sr^{-1}$$ Where $s$ is some symmetry and $r$ is the smallest rotation in fixed direction (more generally it could be arbitrary rotation $r'$ such that $ord(r')=n$). You need to consider three different cases: one when $n$ is odd - every symmetry has precisely one fixed point (vertex), and two cases for even $n$ where symmetry could have either two or zero fixed points.

Immediate consequences are: $$sr^ks^{-1}=r^{-k}$$ $$r^ksr^{-k}=r^{2k}s=sr^{-2k}$$ Note that $D_n=\{s^ir^j:0\leq i \leq 1,0\leq j\leq n-1\}$, so: $$h\in H \implies ({\large\exists_{l \in \mathbb{N}}}:\ h=r^{4l})$$ $${\large \forall_{g\in D_{20}}\forall_{h\in H}\exists_{i,j,l\in \mathbb{N}}}:\ ghg^{-1}=s^ir^jr^{4l}r^{-j}s^{-i}$$ $$s^ir^jr^{4l}r^{-j}s^{-i}=s^ir^{4l}s^{-i}$$ $$s^ir^{4l}s^{-i}\in \{r^{4l}, r^{-4l}\}\subset H$$ $${\large \forall_{g\in D_{20}}\forall_{h\in H}\exists_{i,j,l\in \mathbb{N}}}:\ ghg^{-1}\in H$$ Therefore $H$ is normal subgroup of $G$.

Your sketch of using isomorphic theorem is one way of proving $G/H$ is not abelian. Playing directly on cosets is another option: take $gH, hH\in G/H$. Then $ghH=hgH\iff ghg^{-1}h^{-1}H=1H \iff ghg^{-1}h^{-1} \in H$. Now lets assume that $G/H$ is abelian, in other words: $\forall g,h \in G:\ ghg^{-1}h^{-1} \in H$. Pick $g:=s, h:=r$. Then $ghg^{-1}h^{-1}=srsr^{-1}=r^{-2}$. Because $r^{-2}\in H$ yields contradiction, $G/H$ cannot be abelian.

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