Let Dn = {ar , a; b | 0

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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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The given text appears to be from an abstract algebra or group theory problem focusing on the dihedral group \( D_n \). Here is the transcribed content and explanation suitable for an educational website:

---

### Dihedral Group \( D_n \)

The dihedral group \( D_n \) is defined as:
\[ D_n = \{ a^k, \, a^k b \mid 0 \leq k < n \}, \]
where the elements \( a \) and \( b \) satisfy the following properties:
- \( a^n = e \) (where \( e \) is the identity element),
- \( b^2 = e \),
- \( ba = a^{-1}b \).

Below are the tasks and solutions related to \( D_n \):

#### (a) Show that \((a^k b)^2 = e\) for each \(0 \le k < n\).

To show that \((a^k b)^2 = e\), consider:
\[ (a^k b)^2 = a^k b a^k b. \]

Using the property \( b a = a^{-1} b \), we can rewrite:
\[ a^k b a^k b = a^k (b a^k) b. \]
Because \( b a = a^{-1} b \), we can apply this repeatedly:
\[ b a^k = (b a) a^{k-1} = (a^{-1} b) a^{k-1} = a^{-1} (b a^{k-1}) = a^{-1} (a^{-1} b) a^{k-2} = a^{-2} (b a^{k-2}) = \cdots = a^{-k} b. \]
Thus:
\[ a^k (a^{-k} b) b = a^k a^{-k} (b^2) = e e = e. \]
Therefore:
\[ (a^k b)^2 = e. \]

#### (b) Find the order of each element of \( D_{10} \).

For \( D_{10} \):
- The order of \( e \) is 1.
- The order of \( a^k \) for \( k = 0, 1, ..., 9 \)

Since \( a^{10} = e \), the order of
Transcribed Image Text:The given text appears to be from an abstract algebra or group theory problem focusing on the dihedral group \( D_n \). Here is the transcribed content and explanation suitable for an educational website: --- ### Dihedral Group \( D_n \) The dihedral group \( D_n \) is defined as: \[ D_n = \{ a^k, \, a^k b \mid 0 \leq k < n \}, \] where the elements \( a \) and \( b \) satisfy the following properties: - \( a^n = e \) (where \( e \) is the identity element), - \( b^2 = e \), - \( ba = a^{-1}b \). Below are the tasks and solutions related to \( D_n \): #### (a) Show that \((a^k b)^2 = e\) for each \(0 \le k < n\). To show that \((a^k b)^2 = e\), consider: \[ (a^k b)^2 = a^k b a^k b. \] Using the property \( b a = a^{-1} b \), we can rewrite: \[ a^k b a^k b = a^k (b a^k) b. \] Because \( b a = a^{-1} b \), we can apply this repeatedly: \[ b a^k = (b a) a^{k-1} = (a^{-1} b) a^{k-1} = a^{-1} (b a^{k-1}) = a^{-1} (a^{-1} b) a^{k-2} = a^{-2} (b a^{k-2}) = \cdots = a^{-k} b. \] Thus: \[ a^k (a^{-k} b) b = a^k a^{-k} (b^2) = e e = e. \] Therefore: \[ (a^k b)^2 = e. \] #### (b) Find the order of each element of \( D_{10} \). For \( D_{10} \): - The order of \( e \) is 1. - The order of \( a^k \) for \( k = 0, 1, ..., 9 \) Since \( a^{10} = e \), the order of
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