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Suppose $H \lt D_n$, $|H|=4$

Since $|D_n|=2n$, by Lagrange's Theorem $$4|2n$$ $$2n=4k $$ for some $\in \Bbb Z^+$

$$n=2k$$ thus $n$ is even

Conversely suppose $n$ is even. then how to show $D_n$ has subgroup of order 4 ?

please give me a hint please!

fivestar
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  • When $n$ is even, you can construct a subgroup of order 4 pretty easily. As a hint, one of the elements of the subgroup is going to correspond to a reflection of the rigid body. – Joe May 03 '18 at 15:41
  • Hint. If $n$ is doubly even there's a subgroup of rotations that works. If $n$ singly even, think about the rotation through $\pi$ and the reflection over the $x$-axis. – Ethan Bolker May 03 '18 at 15:41
  • Start with any two distinct reflections that will commute....! – Chinnapparaj R May 03 '18 at 15:43

2 Answers2

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$n$ is even implies $R_{180} \in D_n$ and is of order $2$. So for any reflection $F$, we have $$R_{180}F=FR_{180}$$ and note that any reflection has order $2$ and the latter element has also order $2$.

Hence .................. is your required subgroup!

Chinnapparaj R
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  • ahh! you use $a,b \in G$ and each element of oreder is 2 and commute each other. then ${e,a,b,ab}$ always subgroup of order 4 in $G$ right? – fivestar May 07 '18 at 02:29
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The dihedral groups $D_n$ are supersolvable, hence the converse of Lagrange's Theorem always holds. Since $4\mid D_{2n}$ for even $n\ge 2$, there exists a subgroup of order $4$.

References:

Dihedral group is supersolvable

Complete classification of the groups for which converse of Lagrange's Theorem holds

Dietrich Burde
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