$G=D_6$ and $H=<R^2>$. Use this Cayley table for $D_6$
(a). Show that $H \vartriangleleft G$.
I want to show by finding out $aH=Ha$ for all $a \in G$, but then how do I proceed, it would be too tedious to check all $a$ in $D_6$, is there any way else to show it?
(b). List all cosets of $H$ in $G$.
Since I have $H=<R^2>={I,R^2,R^4}$, should be $IH,RH,R^2H.......F_6H$ and write then all out?
(c). What is the Cayley Table for $G/H$?
What should be the elements of $G/H$? Left cosets of $H$ in $G$?
On
$H=\{id,R^2,R^4\}$ is the unique subgroup of $D_6$ having order $3$. It is a normal subgroup, because we have $R^i(R^2)R^{-1}=R^2, R^i(R^4)R^{-1}=R^4$, and $F_1R^2F_1^{-1}=R^4$, $F_2R^2F_2^{-2}=R^4$ and so on. We just can check this from the Cayley table, that $gHg^{-1}\subseteq H$ for all $g\in D_6$. It follows that the quotient group $D_6/H$ has order $4$. There are only two possible groups of order $4$. In this case we have $D_6/H\simeq C_2\times C_2$, the Kleinian $4$-group, because obviously $D_6/H$ has no element of order $4$.
Let's remember that $$D_6=\langle r,o: o^2=r^6=(or)^2=e \rangle$$ 1. All elements come in the form of $o^m r^n$, with $m=1,0$ we can check that $H$ is normal by showing that $aha^{-1}\in H$, with $a=o^m r^n$ and $h=r^{2s}$ we have $$aha^{-1}=o^m r^n r^{2s} r^{-n} o^{-m}=o^m r^{2s} o^{-m}$$ if $m=0$ or $s=0$ we're done as it is clearly in $H$, if $m=1$ amd s=1 we have $$orro=or(or^{-1})^{-1}=or(or^5)^{-1}=(or)(or)^{-1}r^{-4}=r^2$$ and for $s=2$ $$or^4o=or(or^{-3})^{-1}=or(or^3)^{-1}=(or)(or)^{-1}r^{-2}=r^4$$ Which gives us that $H$ is normal.
The cosets will be $H$ $oH$ and $rH$ because any even power of $o$ will neutralize itself while any even power of $r$ will reduce the power within the normal subgroup by an even number because it goes through all even within it.
Elements in $D_6/H$ are the cosets, so we have $e$, $o$ and $r$ and $or$, which means it's order is 4 and as we have $r^2=o^2=e$ we have that it is isomorphic to $\mathbb{Z}_2\times \mathbb{Z}_2$ under addition.