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I need to be able to find all of the quotient groups for dihedral group 4 with

$D_4 = \{ e,R,R^2,R^3,V,H,D,D'\}$.

I know I have to start by finding the normal subgroups, which are

$\{e,R^2\}$ $\{e,R,R^2,R^3\}$ $\{e,R^2,V,H\}$ $\{e,R^2,D,D'\}$.

Then I need to find the sets defined by $G/H=\{ aH : a \in G\}$ with operation $aH bH = abH$.

I am stuck here; could someone please show me how to find all of the factor groups for D4?

Ralph Clausen
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GroupTheoryHelp
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1 Answers1

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The identity element of the quotient group is the coset which has elements of the normal subgroup. For example $D_4/\{e,R,R^2,R^3\}$ has identity element $\{e,R,R^2,R^3\}$. If the group has order $8$ and the normal subgroup has order $4$, then the order of the quotient group is $2$, so there is only one other element of the quotient group besides the identity -- it's the coset which comes from the group elements not in the normal subgroup.

If the normal subgroup has order $2$, then the quotient subgroup has order $4$. I.e., $D_4/\{e,R^2\}$ has identity $\{e,R^2\}$ and its three cosets $\{R,R^3\}, \{V,H\},$ and $\{D,D'\}.$

It should be noted that you omitted the trivial normal subgroups $\{e\}$ and $D_4.$

J. W. Tanner
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  • So would D4/ {e,R2,V,H} also be {e,R2,V,H} and {e, R, R2, R3}? I think all the order 2 quotient groups are isomorphic – GroupTheoryHelp Jun 20 '19 at 20:55
  • No, {$e,R^2$,V,H} and {$R,R^3,$ D,D'}; cosets are disjoint; yes, all order 2 groups are isomorphic (per my first comment to your question) – J. W. Tanner Jun 20 '19 at 20:56
  • I think I get it now. So would D4/ {e,R2,D,D'} be {e,R2,D,D'} and {R,R3,V,H}? I am looking at the Cayley table to deduce this – GroupTheoryHelp Jun 20 '19 at 21:07
  • It would be that – J. W. Tanner Jun 20 '19 at 21:15
  • I should have mentioned in my answer that the quotient group of order $2$ is isomorphic to the cyclic (or any) group of order $2$, and the quotient group of order $4$ is isomorphic to the Klein four group, in which all three non-identity elements have order $2$, which is not isomorphic to the cyclic group of order $4$ – J. W. Tanner Jun 20 '19 at 21:18