Let n ≥ 1 be a positive integer and Sn the symmetric group on n symbols. Let Δ = {(g, g) ∶ g ∈ Sn}. Which of the following statements are necessarily true?

Question

Let n &ge; 1 be a positive integer and Sn the symmetric group on n symbols. Let &Delta; = {(g, g) &ratio; g &isin; Sn}. Which of the following statements are necessarily true?

Prepp Team · Accepted Answer

Symmetric Group $ S_n $ and Direct Product $ S_n \times S_n $ We are given the symmetric group $ S_n $ on $ n $ symbols, where $ n \ge 1 $ is a positive integer. We are also given the set $ \Delta = \{(g, g) \mid g \in S_n\} $, which is a subset of the direct product $ S_n \times S_n $. The direct product $ S_n \times S_n $ is a group under the operation $ (a_1, b_1)(a_2, b_2) = (a_1 a_2, b_1 b_2) $ for $ (a_1, b_1), (a_2, b_2) \in S_n \times S_n $. The identity element in $ S_n \times S_n $ is $ (e, e) $, where $ e $ is the identity element in $ S_n $. The inverse of an element $ (a, b) \in S_n \times S_n $ is $ (a^{-1}, b^{-1}) $. Analyzing the Statements Map $ f(a, b) = ab $ as a Group Homomorphism Statement 1 claims that the map $ f: S_n \times S_n \to S_n $ given by $ f(a, b) = ab $ is a group homomorphism. For $ f $ to be a homomorphism, it must satisfy $ f((a_1, b_1)(a_2, b_2)) = f(a_1, b_1)f(a_2, b_2) $ for all $ (a_1, b_1), (a_2, b_2) \in S_n \times S_n $. Let's evaluate both sides: Left side: $ f((a_1, b_1)(a_2, b_2)) = f(a_1 a_2, b_1 b_2) = (a_1 a_2)(b_1 b_2) $. Right side: $ f(a_1, b_1)f(a_2, b_2) = (a_1 b_1)(a_2 b_2) $. So, for $ f $ to be a homomorphism, we must have $ (a_1 a_2)(b_1 b_2) = (a_1 b_1)(a_2 b_2) $ for all $ a_1, b_1, a_2, b_2 \in S_n $. Let $ a_2 = e $ (the identity element in $ S_n $). Then the condition becomes $ (a_1 e)(b_1 b_2) = (a_1 b_1)(e b_2) $, which simplifies to $ a_1 (b_1 b_2) = (a_1 b_1) b_2 $. This is true by the associativity of the group operation in $ S_n $. However, the condition must hold for all $ a_1, b_1, a_2, b_2 $. Let's rearrange the required equality $ a_1 a_2 b_1 b_2 = a_1 b_1 a_2 b_2 $. Multiplying by $ a_1^{-1} $ on the left and $ b_2^{-1} $ on the right gives $ a_2 b_1 = b_1 a_2 $. This must hold for all $ a_2, b_1 \in S_n $. This means that every element in $ S_n $ must commute with every other element in $ S_n $. In other words, $ S_n $ must be an abelian group. The symmetric group $ S_n $ is abelian only for $ n=1 $ (where $ S_1 $ has one element) and $ n=2 $ (where $ S_2 $ has two elements and is isomorphic to $ \mathbb{Z}_2 $). For $ n \ge 3 $, $ S_n $ is not abelian. Since the condition $ a_2 b_1 = b_1 a_2 $ is not generally true for all elements in $ S_n $ when $ n \ge 3 $, the map $ f(a, b) = ab $ is not necessarily a group homomorphism for all $ n \ge 1 $. &Delta; as a Subgroup of $ S_n \times S_n $ Statement 2 claims that $ \Delta $ is a subgroup of $ S_n \times S_n $. To verify this, we check the subgroup criteria: Identity Element: The identity element of $ S_n \times S_n $ is $ (e, e) $, where $ e $ is the identity in $ S_n $. Since $ e \in S_n $, the element $ (e, e) $ is in $ \Delta $ by definition of $ \Delta $. Closure: Let $ x = (g_1, g_1) $ and $ y = (g_2, g_2) $ be two arbitrary elements in $ \Delta $. By the group operation in $ S_n \times S_n $, their product is $ xy = (g_1, g_1)(g_2, g_2) = (g_1 g_2, g_1 g_2) $. Since $ g_1 \in S_n $ and $ g_2 \in S_n $, their product $ g_1 g_2 $ is also in $ S_n $ (because $ S_n $ is a group). Therefore, the element $ (g_1 g_2, g_1 g_2) $ is of the form $ (h, h) $ where $ h = g_1 g_2 \in S_n $, which means it is in $ \Delta $. $ \Delta $ is closed under the group operation. Inverses: Let $ x = (g, g) $ be an arbitrary element in $ \Delta $. The inverse of $ x $ in $ S_n \times S_n $ is $ x^{-1} = (g^{-1}, g^{-1}) $. Since $ g \in S_n $, its inverse $ g^{-1} $ is also in $ S_n $ (because $ S_n $ is a group). Therefore, the element $ (g^{-1}, g^{-1}) $ is of the form $ (h, h) $ where $ h = g^{-1} \in S_n $, which means it is in $ \Delta $. Every element in $ \Delta $ has its inverse in $ \Delta $. Since $ \Delta $ satisfies all three subgroup criteria, $ \Delta $ is a subgroup of $ S_n \times S_n $ for all $ n \ge 1 $. &Delta; as a Normal Subgroup of $ S_n \times S_n $ Statement 3 claims that $ \Delta $ is a normal subgroup of $ S_n \times S_n $. A subgroup $ H $ of a group $ G $ is normal if for every $ x \in G $ and $ h \in H $, the element $ xhx^{-1} $ is in $ H $. Here, $ G = S_n \times S_n $ and $ H = \Delta $. Let $ x = (a, b) \in S_n \times S_n $ and $ h = (g, g) \in \Delta $. The inverse of $ x $ is $ x^{-1} = (a^{-1}, b^{-1}) $. We compute $ xhx^{-1} $: $ xhx^{-1} = (a, b)(g, g)(a^{-1}, b^{-1}) = (ag, bg)(a^{-1}, b^{-1}) = (ag a^{-1}, bg b^{-1}) $ For $ \Delta $ to be a normal subgroup, $ (ag a^{-1}, bg b^{-1}) $ must be in $ \Delta $ for all $ (a, b) \in S_n \times S_n $ and all $ (g, g) \in \Delta $. This means we must have $ ag a^{-1} = bg b^{-1} $ for all $ a, b, g \in S_n $. Let's test this condition. Let $ g \in S_n $ be any element. Take $ a = e $ (the identity) and $ b $ to be any element in $ S_n $. The condition becomes $ eg e^{-1} = bg b^{-1} $, which simplifies to $ g = bg b^{-1} $. This implies that $ g $ must commute with every element $ b \in S_n $ (i.e., $ gb = bg $). This must hold for all $ g \in S_n $. This would mean that $ S_n $ must be an abelian group. As discussed earlier, $ S_n $ is abelian only for $ n=1 $ and $ n=2 $. For $ n \ge 3 $, $ S_n $ is not abelian. Therefore, for $ n \ge 3 $, there exist elements $ g, b \in S_n $ such that $ gb \ne bg $, which means $ g \ne bg b^{-1} $. In this case, taking $ a=e $, the element $ (ag a^{-1}, bg b^{-1}) = (g, bg b^{-1}) $ is not in $ \Delta $ unless $ g = bg b^{-1} $. For example, in $ S_3 $ ($ n=3 $), let $ g = (1 2) $, $ a = e $, $ b = (1 3) $. Then $ ag a^{-1} = (1 2) $ and $ bg b^{-1} = (1 3)(1 2)(1 3)^{-1} = (1 3 2) $. Since $ (1 2) \ne (1 3 2) $, the element $ ((1 2), (1 3 2)) $ is not in $ \Delta $. Thus, $ \Delta $ is not a normal subgroup of $ S_3 \times S_3 $. Since $ \Delta $ is not normal for all $ n \ge 1 $ (specifically, not for $ n \ge 3 $), statement 3 is not necessarily true. &Delta; as a Normal Subgroup if $ n $ is Prime Statement 4 claims that $ \Delta $ is a normal subgroup of $ S_n \times S_n $ if $ n $ is a prime number. The condition for normality is still $ ag a^{-1} = bg b^{-1} $ for all $ a, b, g \in S_n $. As shown above, this requires $ S_n $ to be abelian, which is only true for $ n=2 $. If $ n $ is a prime number, it doesn't automatically make $ S_n $ abelian. For instance, if $ n=3 $ (which is prime), $ S_3 $ is not abelian, and as shown earlier, $ \Delta $ is not a normal subgroup of $ S_3 \times S_3 $. Therefore, the primality of $ n $ does not ensure that $ \Delta $ is a normal subgroup. Statement 4 is not necessarily true. Conclusion Based on the analysis: Statement 1 is false for $ n \ge 3 $. Statement 2 is true for all $ n \ge 1 $. Statement 3 is false for $ n \ge 3 $. Statement 4 is false for prime $ n \ge 3 $. The only statement that is necessarily true for all $ n \ge 1 $ is that $ \Delta $ is a subgroup of $ S_n \times S_n $.

Prepp Team · Answer

The map f &ratio; S n  &times; S n  &rarr; S n  given by f(a, b) = ab is a group homomorphism.

Prepp Team · Answer

&Delta; is a normal subgroup of S n  &times; S n .

Prepp Team · Answer

&Delta; is a normal subgroup of S n  &times; S n , if n is a prime number.

Symmetric Group \( S_n \) and Direct Product \( S_n \times S_n \)

Analyzing the Statements

Map \( f(a, b) = ab \) as a Group Homomorphism

Δ as a Subgroup of \( S_n \times S_n \)

Δ as a Normal Subgroup of \( S_n \times S_n \)

Δ as a Normal Subgroup if \( n \) is Prime

Conclusion

Important Questions from Group & Subgroups

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