Let's solve each part step by step, using the definition of **internal product** and the fact that if a group \( G \) is the internal product of two normal subgroups \( N_1 \) and \( N_2 \), then \( G \cong N_1 \times N_2 \), provided \( N_1 \cap N_2 = \{e\} \). --- ## 1. Find an isomorphic group to \( D_{2n} \times \mathbb{Z}_2 \) and identify the two normal subgroups \( N_1 \) and \( N_2 \). ### **Step 1: Recall definitions** - \( D_{2n} \): Dihedral group of order \( 2n \), generated by \( r \) (rotation) and \( s \) (reflection), with the relations: \( r^n = e,\, s^2 = e,\, srs^{-1} = r^{-1} \). - \( \mathbb{Z}_2 \): Cyclic group of order 2. ### **Step 2: Structure of \( D_{2n} \times \mathbb{Z}_2 \)** Let \( G = D_{2n} \times \mathbb{Z}_2 \). - \( D_{2n} = \langle r, s \rangle \) - \( \mathbb{Z}_2 = \langle t \rangle \), \( t^2 = e \). ### **Step 3: Identify possible normal subgroups** Let: - \( N_1 = D_{2n} \times \{e\} \) - \( N_2 = \{e\} \times \mathbb{Z}_2 \) Both are normal (since they're direct factors), and \( N_1 \cap N_2 = \{(e,e)\} \). Moreover, - \( N_1 \cong D_{2n} \) - \( N_2 \cong \mathbb{Z}_2 \) ### **Step 4: Isomorphism** By the internal product theorem, since \( G = N_1N_2 \) and \( N_1 \cap N_2 = \{e\} \), \( G \cong N_1 \times N_2 = D_{2n} \times \mathbb{Z}_2 \). Thus, **an isomorphic group is \( D_{2n} \times \mathbb{Z}_2 \) itself**, with the two normal subgroups: - \( N_1 = D_{2n} \times \{e\} \) - \( N_2 = \{e\} \times \mathbb{Z}_2 \) > **Conclusion:** > \( D_{2n} \times \mathbb{Z}_2 \cong N_1 \times N_2 \), with > \( N_1 = D_{2n} \times \{e\} \), \( N_2 = \{e\} \times \mathbb{Z}_2 \). --- ## 2. Show that \( D_{4n} \) is **not** isomorphic to \( D_{2n} \times \mathbb{Z}_2 \). Take \( n=2 \) and **list all elements of order 2** in \( D_8 \) and \( D_4 \times \mathbb{Z}_2 \), and the generators if any for both. ### **Step 1: Orders and Elements** - \( D_8 \): Dihedral group of order 8 (\( n=4 \)), elements: \( \{e, r, r^2, r^3, s, sr, sr^2, sr^3\} \), where \( r^4 = e, s^2 = e, srs = r^{-1} \). - \( D_4 \): Dihedral group of order 4 (\( n=2 \)), elements: \( \{e, r, s, sr\} \) with \( r^2 = e, s^2 = e \). - \( D_4 \times \mathbb{Z}_2 \): order 8. ### **Step 2: List elements of order 2 in \( D_8 \)** Let's compute the orders: - \( r \): order 4 - \( r^2 \): order 2 - \( r^3 \): order 4 - \( s \): order 2 - \( sr \): order 2 - \( sr^2 \): order 2 - \( sr^3 \): order 2 So, elements of order 2 in \( D_8 \): - \( r^2 \) - \( s \) - \( sr \) - \( sr^2 \) - \( sr^3 \) Total: **5 elements** of order 2. #### **Check:** - \( (sr^k)^2 = s r^k s r^k = s s r^{-k} r^k = e \), so all \( sr^k \) have order 2. - \( r^2 \) has order 2. ### **Step 3: List elements of order 2 in \( D_4 \times \mathbb{Z}_2 \)** - \( D_4 \) elements: \( e, r, s, sr \) (\( r^2 = e, s^2 = e \)) - \( \mathbb{Z}_2 \): \( e, t \) (\( t^2 = e \)) All elements: \( (x, y) \) Compute order for each: 1. \( (e, e) \): order 1 2. \( (e, t) \): order 2 3. \( (r, e) \): order 2 4. \( (r, t) \): order 2 5. \( (s, e) \): order 2 6. \( (s, t) \): order 2 7. \( (sr, e) \): order 2 8. \( (sr, t) \): order 2 Check: - \( (r, e)^2 = (r^2, e^2) = (e, e) \) - \( (r, t)^2 = (r^2, t^2) = (e, e) \) - Similarly for others. **All elements except \( (e, e) \) have order 2.** So, **7 elements of order 2** in \( D_4 \times \mathbb{Z}_2 \). ### **Step 4: Generators** - \( D_8 \): \( r \) and \( s \) generate the group, since \( r^4=e, s^2=e, srs = r^{-1} \). - \( D_4 \times \mathbb{Z}_2 \): None of the elements have order 4, so cannot generate the group with a single element or with a pair that gives an element of order 4. ### **Step 5: Compare** - In \( D_8 \), **5 elements** of order 2. - In \( D_4 \times \mathbb{Z}_2 \), **7 elements** of order 2. > **Conclusion:** > Since the number of elements of order 2 differs between the two groups, \( D_8 \not\cong D_4 \times \mathbb{Z}_2 \). > More generally, \( D_{4n} \not\cong D_{2n} \times \mathbb{Z}_2 \). --- ## 3. Find an isomorphism \( N_1 \times N_2 \cong U_9 \) (the multiplicative group of units mod 9). ### **Step 1: Structure of \( U_9 \)** \( U_9 = \{1,2,4,5,7,8\} \) (numbers less than 9 and coprime to 9). Order is 6. Let's find the group structure: - \( 2^1 = 2 \) - \( 2^2 = 4 \) - \( 2^3 = 8 \) - \( 2^4 = 7 \) - \( 2^5 = 5 \) - \( 2^6 = 1 \) So \( 2 \) is of order 6, so \( U_9 \) is **cyclic** of order 6. ### **Step 2: Subgroups** - Cyclic group of order 6 is isomorphic to \( \mathbb{Z}_6 \cong \mathbb{Z}_2 \times \mathbb{Z}_3 \). - The only normal subgroups are those of order 2 and 3. Let - \( N_1 = \langle 4 \rangle = \{1, 4, 7\} \) (order 3). - \( N_2 = \langle 8 \rangle = \{1, 8\} \) (order 2). -