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I ran into this question when I was studying for my abstract algebra midterm.

Show that the subgroup $H$ of rotations is normal in the dihedral group $D_n$. Find the quotient group $D_n/H$.

I'm not quite sure where to begin. I know that for a Dihedral group of $n\geq 3$, then $r^n=1$ where $r$ is a rotation, and $s^2=1$ where $s$ is a reflection, and $srs=r^{-1}$. I was not sure how to prove something is a normal subgroup from here. Any advice, thanks!

HK Lee
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Richard Cao
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4 Answers4

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The index $2$ suggestion works, but you can also show this directly. One can check that the generators $R$ and $F$ of the dihedral group conform to the rule $RF = FR^{-1}$. From this, we see that any element in $D_n$ can be written as $R^jF^k$ where $0 \leq j \leq n-1$ and $0 \leq k \leq 1$.

A subgroup $N \leq G$ is normal whenever, given any $n \in N$ and $g \in G$, we have $gng^{-1} \in N$. In this case, any element of the rotation subgroup looks like $R^m$ for $1 \leq m \leq n-1$. Considering any arbitrary element $R^jF^k$ of $D_n$, we just need to show that $(R^jF^k)R^m(R^jF^k)^{-1} \in \langle R \rangle$. Clearly this is true if $k=0$, so assume $k=1$. Now look to the helpful rule in the first paragraph to conclude that this is indeed an element of $\langle R \rangle$.

Kaj Hansen
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If $D_n=\langle r,s\mid r^n=s^2=1,srs=r^{-1}\rangle$, then $D_n$ has order $2n$ and the group generated by $r$ has order $n$.

Therefore the index of $\langle r\rangle$ in $D_n$ is equal to two, and it is a general fact that if $H\leq G$ is a subgroup with $[G:H]=2$ then $H$ is a normal subgroup of $G$.

carmichael561
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  • Without knowing that a subgroup of index 2 is normal, is there any other way to solve this problem, or will such a solution be extremely unwieldy? Thanks again! – Richard Cao Mar 27 '17 at 20:24
  • I suppose you could try to verify it directly, but this would likely amount to the proof that every subgroup of index $2$ is normal. – carmichael561 Mar 27 '17 at 20:26
  • I see. Thanks again! – Richard Cao Mar 27 '17 at 20:29
  • @Richard You could just directly conjugate a general rotation by a general reflection and observe that the result is a rotation. But that doesn't sound like fun. – Matt Samuel Mar 27 '17 at 20:36
  • @MattSamuel It suffices to consider any one reflection, e.g. $s$, and show that $sr^ks$ is a rotation for any $k$. This shows that the normalizer of $\langle r \rangle$ contains an element not in $\langle r \rangle$, and therefore (by order considerations) must be all of $D_n$. –  Mar 28 '17 at 06:26
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How many rotations are there, and how does this compare to the total number of elements? You may have shown as an exercise previously that a subgroup of index $2$ is normal. That is relevant here. If not, you should prove it, because using this fact is the easiest way I see to solve your problem.

Matt Samuel
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This may not count as a proof (depending on your definition of the dihedral group) but it explains geometrically what's going on. The dihedral group is the group of symmetries of the regular $n$-gon in the plane. It consists of $n$ rotations, which clearly form a subgroup, and $n$ reflections. If you think geometrically, the product of a rotation $r$ and a reflection $s$ reverses orientation in the plane, so must be a reflection.

Note that $grg^{-1}$ is clearly a rotation if $g$ is. It is a product of two reflections if $g$ is a reflection, so preserves orientation and must be a rotation. That says the subgroup of rotations is invariant under conjugation, so normal.

Ethan Bolker
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  • It does not take much more to turn this into a completely rigorous proof (and this is my preferred proof, because it explains the geometric origins of the dihedral group). Namely, one verifies that the dihedral group is isomorphic to the finite subgroup $G < O(2)$ as described in this answer. Then one verifies with a simple matrix computation that the rotations of $O(2)$ form a normal subgroup of $O(2)$. Since the intersection of $G<O(2)$ with any normal subgroup of $O(2)$ is a normal subgroup of $G$, the proof is done. – Lee Mosher May 04 '18 at 14:14
  • And of course, as you say, perhaps this is one's definition of the dihedral group (also my preferred definition). – Lee Mosher May 04 '18 at 14:25