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Using this operations table i'm trying to figure out how to find the order of the factor group $D_{6} / H$ with H = $ \{ \rho_{0} , \rho_{3} \}$ and then figure out which well-known group it's isomorphic to. I know the order of $D_{6}$ on it's own is 12 and I know how to find the order of a factor group like $( \Bbb Z_{4} \times \Bbb Z_{2}) \space / \space \langle (2,1) \rangle$, but I can't figure out how to calculate this factor group. Maybe once I can figure it out, then I might be able to find what it's isomorphic to? Or are those completely unrelated. Any advice would be greatly appreciated.

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Cosets of a normal subgroup are pairwise disjoint and cover all of the group, so for finite groups we get $|G/H|=|G|/|H|$. This is usually very helpful in classifying factor groups of finite groups, and your problem is no exception.

As to which order-6 group $D_6/H$ is, we only have two options: $\Bbb Z_6$ or $S_3$.

If it is to be $\Bbb Z_6$, then there must be an order 6 element in $D_6/H$. An order 6 element in the factor group must come from an element in the original group that has order which is a multiple of 6. There are only two such: $\rho_1$ and $\rho_5$. However, $\rho_1H$ and $\rho_5H$ have order $3$ as $\rho_1^3=\rho_5^3=\rho_3\in H$, so the group cannot be $\Bbb Z_6$.

If it is to be $S_3$, there must be two non-commuting elements in $D_6/H$. Non-commuting elements in the factor group must come from non-commuting elements in the original group. We try with $\rho_1$ and $\mu_1$, because those are the first two non-commuting elements in the table: $$ \rho_1H\cdot\mu_1H=(\rho_1\mu_1)H=\delta_1H=\{\delta_1,\mu_3\}\\ \mu_1H\cdot\rho_1H=(\mu_1\rho_1)H=\delta_3H=\{\delta_3,\mu_2\} $$ These are not the same coset, so $\rho_1H$ and $\mu_1H$ do not commute. So the group must be $S_3$.

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  • $\begingroup$ Thank you for clearing this up for me, just a couple more questions if you don't mind. How did you get that $\rho_{1}$ and $\rho_{5}$ were multiples of 6 and also how did you get that they are both order 3? $\endgroup$ – Matt L. Jul 13 at 6:40
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    $\begingroup$ @MattL. The $\mu_i$ and the $\delta_i$ so have order $2$, so those cannot be considered. To check the $\rho_i$, the subgroup consisting of only $\rho_i$'s is isomorphic to $\Bbb Z_6$ by way of $\rho_i\mapsto i$. And in $\Bbb Z_6$, the only elements of order $6$ are $1$ and $5$. So $\rho_1$ and $\rho_5$ are the only elements in $D_6$ of order $6$ (and no element has larger order). As for the orders of $\rho_1H$ and $\rho_5H$, note that $$(\rho_1H)^3=\rho_1^3H=\rho_3H=H$$ (and similarly for $\rho_5$), so in the factor group the elements $\rho_1H$ and $\rho_5H$ have order 3 rather than order 6. $\endgroup$ – Arthur Jul 13 at 8:53
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    $\begingroup$ (In the factor group, $H=\rho_0H=\rho_3H$ is the identity element.) $\endgroup$ – Arthur Jul 13 at 8:54
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    $\begingroup$ @MattL. I am not entirely certain what you mean. For two cosets to be the same, they have to have the same elements, yes. And two elements $x,y$ being in the same coset of $H$ means that $xH=yH$. Does that clear things up? $\endgroup$ – Arthur Jul 13 at 23:10
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    $\begingroup$ Yes. Cosets are, ultimately, just subsets of the group, and equality of cosets come from there. $\endgroup$ – Arthur Jul 13 at 23:20
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Hope you know the following easy facts( even if you are unaware, you can directly prove, it would be worth exercises):

  1. The only abelian group of order 6 is the cyclic group namely $\mathbb{Z}_{6}$.
  2. If $H$ is any subgroup of $G$ such that $H \subseteq Z(G)$, and if $G/H$ is cyclic then $G$ is abelian.
  3. The only non abelian group of order 6 is $S_{3}$.

Now come to your question. Hope you are convinced that this factor group is of order 6 and observe from this table that $D_{6}$ is non abelian and $H$ lies in the center. Hence from the above fact, the only possibility of this factor group is $S_{3}$.

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As $|D_6|=12$ and $|H|=2$, the quotient group $D_6/H$ will have order $12/2=6$.

The group $D_6$ may be written out $$\{1,x,x^2,x^3,x^4,x^5, y,yx,yx^2,yx^3,yx^4,yx^5\}$$ where $x,y\in D_6$ represent a rotation of $\mathbb{R}^2$ through $60^\circ$ and a reflection of $\mathbb{R}^2$, about a line through the origin.

The group is characterised by the identities, $x^6=y^2=1, yxy^{-1}=x^{-1}$. To see this just note that the product of any two elements in the list may be reduced to another element in the list using just these identities.

To describe the quotient group, we only need to add the relation that $x^3=1$ (as $H=\{1,x^3\}$). However we may now throw away the relation that $x^6=1$ as it is implied by $x^3=1$. That is in $D_6/H$ we have $$x^3=1, y^2=1, yxy^{-1}=x^{-1}.$$

Our elements are now: $$D_6/H=\{1,x,x^2, y,yx,yx^2\}$$ Note these must be distinct as we must have $6$ elements. Given the relations these elements satisfy, we may recognise this group as $D_3$.

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