I want to find $N(D_{4})/D_{4}$ where $N(D_{4})$ is the normalizer of $D_{4}$ in $D_{16}$. I'm not too clear on what the normalizer of $D_{4}$ in $D_{16}$

Is there a nice way to find $N(D_{4})/D_{4}$? I can't seem to figure this exercise out and would appreciate someone explaining it to me.


This is a situation where explicit calculations are the way to go. This will often work even when there is a better way and often leaves you with a better understanding of what's going on (I know I understand the dihedral group better after doing this), so you may want to try it first before trying something really clever.

Let $r,s$ be the generating rotation and reflection of $D_{16}$, respectively. Then $D_4$ is the subgroup generated by $r^4$, $s$. If $t$ normalizes $D_4$, then $$tr^4t^{-1}=r^{4a}$$ for some $a$ (since any inner automorphism sends rotations to rotations and reflections to reflections), and $$tst^{-1}=r^{4b}s$$ for some $b$. Thus $$tr^4st^{-1}=r^{4(a+b)}s$$ Let $t=r^xs^y$ where $0\leq x<16$ and $y\in \{0,1\}$.

Suppose first that $y=0$. Then $$tr^4t^{-1}=r^4$$ and $$tst^{-1}=r^xsr^{-x}=r^{2x}s$$ Thus $x$ must be a multiple of $2$ in order for $t$ of this form to normalize $D_4$.

Suppose next that $y=1$. Then $$r^xsr^4sr^{-x}=r^{-4}$$ and $$r^xsssr^{-x}=r^{2x}s$$ Thus $x$ being a multiple of $2$ is also the only possibility in this case. We can conclude that the normalizer is generated by $r^2,s$ and hence isomorphic to $D_8$. Since this has twice as many elements as $D_4$, your quotient is a group of order $2$.

  • $\begingroup$ Thank you for the time for such a well thought out answer. This really helped me understand better. $\endgroup$ – LonrAbstract Dec 16 '15 at 5:25
  • $\begingroup$ @LonrAbstract no problem. $\endgroup$ – Matt Samuel Dec 16 '15 at 5:25

A slightly faster way to get to the same place:

Since $rs(r^4s)rs = r(r^{12}s)srs = (r^{13})(rs) = r^{14}s \not\in D_4$, it follows that:

$N(D_4) \neq D_{16}$.

On the other hand, $D_8 = \langle r^2,s\rangle$ contains $D_4 = \langle r^4, s\rangle$, and $D_4$ has index $2$ in this subgroup of $D_{16}$, and is thus necessarily normal (this also eliminates the possibility that $N(D_4) = D_4$).

Since $D_8$ (in this form) is a maximal subgroup of $D_{16}$, it must be the normalizer, and as we have seen from the index, the quotient has order $2$.

(This uses the fact that the normalizer of a subgroup $H$ in a group $G$ is the largest subgroup of $G$ containing $H$ in which $H$ is normal).

If one looks closely, one see that this is essentially an "order argument", and only one conjugation is calculated. Since $[D_{16}:D_4] = 4$, the only possibilities for the normalizer's order is $8,16$ or $32$.

(Almost irrelevant comment: powers of two play an essential role in the "character" of dihedral groups).

  • $\begingroup$ These are both really great answers. Thank you! $\endgroup$ – LonrAbstract Dec 16 '15 at 5:24

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