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Let $n,m$ be integers with $n \ge 3$ and $m \ge 1$ and where $m$ divides $n$.

Prove that the dihedral group of order $2n$, denoted $D_n$ (instead of $D_{2n}$), has subgroup of order $m$ and then of order $2m$.

We have

$$D_n := \langle r,s \rangle = \{1_{D_n},r,r^2, ..., r^{n-1},$$

$$s, rs, r^2s, ..., r^{n-1}s\},$$

where $order(r)=n$ and $order(s)=2$ and $sr=r^{-1}s$.

Questions:

  1. Prove it has a subgroup of order $m$: (move to answer)

  2. Prove it has a subgroup of order $2m$: Well I guess we can't get away with cyclic again. I'll try $\langle s, r^k\rangle$. Does this work? Proof outline in answer. Is it right?

  3. This $2m$ subgroup I've chosen is $D_m$ right? (yes. i'll just say in answer.)

BCLC
  • 13,459

2 Answers2

1

There are such general simple facts.

If $H$ and $F$ are subgroups of group $G$ and $H$ is normal in $G$, then $HF$ is a subgroup of $G$. If additionally the group $H$ and $F$ are finite, then $|HF|=|H|\cdot|F|/|H\cap F|$.

In our case $H=\langle r^k\rangle$ is a normal subgroup $D_n$ (since $srs^{-1}=r^{-1}$) and $\langle r^k\rangle\cap\langle s\rangle=\{e\}$ for any $k$. Hence $HS$ ($S=\langle s\rangle$) is a subgroup of $D_n$ and if $k$ is a divisor of $n$ and $m=n/k$, then $|HS|=2m$.

kabenyuk
  • 10,712
0

For 1: For $\frac{n}{m}=k$, it's just $\langle r^{k}\rangle$ because $order(r^{k})=\frac{order(r)}{\gcd(order(r),k)} = \frac{n}{\gcd(n,k)} = \frac{n}{k} = m$.

Proof outline for (2): (I can type up the full details if you want, but I'm not really sure anyone will really verify, unless I make another post or something, if I do anyway.)

I claim the elements are $\langle s, r^k\rangle = A$, where

$$A := \{1_{D_n},r^k,r^{2k}, ..., r^{(m-1)k},$$

$$s, rs, r^ks, r^{2k}s, ..., r^{(m-1)k} s\}.$$

I believe this is true if and only if for any $i,j = 0, 1, ..., m-1$ that $A$ is indeed closed and has $2m$ distinct elements

2.1. Closure Part 1: $r^{ik}r^{jk} \in A$,

2.2. Closure Part 2: $sr^{ik}sr^{jk} \in A$

2.3. Closure Part 3: $sr^{ik}r^{jk} \in A$

2.4. Distinct Part 1: $r^{ik} \ne r^{jk}$ if $i \ne j$

2.5. Distinct Part 2: $sr^{ik} \ne r^{jk}$ if $i \ne j$

2.6. Distinct Part 3: $sr^{ik} \ne sr^{jk}$ if $i \ne j$

QED

For 3:

Yes, it's $D_m$. Our new $(s,r)$ is $(s,r^k)$.

BCLC
  • 13,459
  • 1
    There are such general simple facts. If $H$ and $F$ are subgroups of group $G$ and $H$ is normal in $G$, then $HF$ is a subgroup of $G$. If additionally the group $H$ and $F$ are finite, then $|HF|=|H|\cdot|F|/|H\cap F|$. In your case $\langle r^k\rangle$ is a normal subgroup $D_n$ for any $r$ and $\langle r^k\rangle\cap\langle s\rangle={e}$. – kabenyuk Oct 28 '21 at 02:51