MATH 351 – Modern Algebra 1 Selected Solutions – Problem Set 8 November 10, 2018

5.10. We are given that N_{1}, …, N_{k} are finite normal subgroups of some group G such that gcd(|N_{i}|, |N_{j}|) = 1 whenever i ≠ j and we want to show that

N_{1} × ⋯ × N_{k} ≅ N_{1}N_{2} ⋯ N_{k}.

It is true that this is intimately connected with Lemmas 5.1 and 5.2 and Theorem 5.2. However there are some differences, too, that we must take into account. First, note that we are not assuming G = N_{1} ⋯ N_{k}, so it might not be true that G is the internal direct product of the subgroups N_{i}. The product N_{1} ⋯ N_{k} could be some proper subgroup of G. Second, the hypothesis gcd(|N_{i}|, |N_{j}|) = 1 here is different from the condition that appears in the definition of the internal direct product. Hence there are several different ways to proceed here.

Solution 1: (The “slickest” way.) We argue by induction on k. If k = 1, there is nothing to show. If this seems unsatisfying as a base case, consider the case k = 2. Then note that gcd(|N_{1}|, |N_{2}|) = 1 implies that N_{1} ∩ N_{2} = {e} by Lagrange’s theorem, since N_{1} ∩ N_{2} is a subgroup of both N_{1} and N_{2}. Then N_{1}×N_{2} ≅ N_{1}N_{2} = H follows from Theorem 5.2 applied to the group H = N_{1}N_{2} (not G), since H is the internal direct product of its subgroups N_{1} and N_{2}. Now assume the statement holds for all collections of k subgroups satisfying the given conditions, and consider N_{1}, …, N_{k}, N_{k+1}. By the induction hypothesis, N_{1}N_{2} ⋯ N_{k} is isomorphic to N_{1} × N_{2} × ⋯ × N_{k}. In particular,

|N_{1}N_{2} ⋯ N_{k}| = |N_{1}||N_{2}| ⋯ |N_{k}|.

It follows that gcd(|N_{1}N_{2} ⋯ N_{k}|, |N_{k+1}|) = 1, since gcd(|N_{i}|, |N_{k+1}|) = 1 for all 1 ≤ i ≤ k. But this implies

(N_{1}N_{2} ⋯ N_{k}) ∩ N_{k+1} = {e},

as above. Since N_{1}N_{2} ⋯ N_{k} is normal in G, the proof that

(N_{1} × N_{2} × ⋯ × N_{k}) × N_{k+1} ≅ (N_{1}N_{2} ⋯ N_{k})N_{k+1}

will follow exactly as in the case k = 2 done above.

Solution 2: (This one is essentially repeating the proof of Theorem 5.2, which is OK, but less “elegant.”) Consider the mapping

\begin{matrix}α : N_{1} × N_{2} × ⋯ × N_{k} & → N_{1} ⋯ N_{k} \\ (n_{1}, n_{2}, …, n_{k}) & ↦ n_{1}n_{2} ⋯ n_{k} \end{matrix}

Then • α is surjective by definition.

• α is a group homomorphism by Lemma 5.1 and the same argument as in the text in the proof of Theorem 5.2:

α(n_{1}n′_{1}, n_{2}n′_{2}, …, n_{k}n′_{k}) = (n_{1}n′_{1})(n_{2}n′_{2}) ⋯ (n_{k}n′_{k})
= (n_{1}n_{2} ⋯ n_{k})(n′_{1}n′_{2} ⋯ n′_{k})
= α(n_{1}, n_{2}, …, n_{k})α(n′_{1}, n′_{2}, …, n′_{k}),

since Lemma 5.1 implies that the n′_{i} and the n_{j} all commute with each other.

• Finally, we show that α is injective, using induction on k. If k = 1, there is nothing to prove. So assume that the result is true whenever k = ℓ and consider products of ℓ + 1 subgroups. To show α is injective, it suffices to show that ker(α) = (e, e, …, e). So consider (n_{1}, n_{2}, …, n_{ℓ+1}) such that

α(n_{1}, n_{2}, …, n_{ℓ}, n_{ℓ+1}) = n_{1}n_{2} ⋯ n_{ℓ}n_{ℓ+1} = e.

This implies

n_{ℓ+1}^{−1} = n_{1}n_{2} ⋯ n_{ℓ} ∈ (N_{1}N_{2} ⋯ N_{ℓ}) ∩ N_{ℓ+1}.

In the “slick” solution, we saw that (N_{1}N_{2} ⋯ N_{ℓ}) ∩ N_{ℓ+1} = {e} because N_{1}N_{2} ⋯ N_{ℓ} is isomorphic to N_{1} × N_{2} × ⋯ × N_{ℓ} so the order of N_{1}N_{2} ⋯ N_{ℓ} is the product of the orders of the N_{i}, hence is relatively prime to |N_{ℓ+1}|. The same is true here. Hence that implies n_{ℓ+1} = e and then by induction n_{1}n_{2} ⋯ n_{ℓ} = e implies n_{1} = n_{2} = ⋯ = n_{ℓ} = e as well.

5.18. In the cyclic group G = ⟨a⟩ of order 90 = 2 · 3^{2} · 5, the subgroup of 2-elements is H_{2} = {e, a^{45}} = ⟨a^{45}⟩. The subgroup of 3-elements is ⟨a^{10}⟩. (Note that a^{10}, a^{20}, a^{40}, a^{50}, a^{70}, a^{80} have order 9, while a^{30} and a^{60} have order 3, and e has order 1. So all the elements of ⟨a^{10}⟩ are 3-elements and they the only ones.) Similarly, the subgroup of 5-elements is ⟨a^{18}⟩. Since G is the internal direct product of its subgroups H_{2}, H_{3}, H_{5} by Lemma 5.5, we know there is exactly one way to write a = bcd with b ∈ H_{2}, c ∈ H_{3} and d ∈ H_{5}. A little experimentation gives

b = a^{45}, c = a^{10}, d = a^{36}.

(Note, it would also be possible to solve this by setting up a congruence on the exponents writing b = a^{45i}, c = a^{10j}, d = a^{18k}. We want to find i, j, k with 45i+10j +18k ≡ 1 mod 90, and i = 1, j = 1, k = 2 is clearly one solution.)