Algebra Solutions: Hungerford I.8

1) (Qual 2001 #1) S_3 is not the direct product of any family of its proper subgroups. The same is true of \mathbb{Z}_{p^n} and \mathbb{Z}
Note that the order of S_3 is 6, so any subgroup must be of order 2 or 3, which are prime numbers. Any group of prime order must be cyclic, and cyclic groups are abelian. The direct product of abelian groups is abelian, but S_3 is not abelian so it cannot be ismorphic to an abelian group, and hence is not isomorphic to a direct product of proper subgroups.
Now, let p be prime and n \in \mathbb{N}. We have \mathbb{Z}_{p^n} is cyclic of order p^n. Thus any proper subgroup of \mathbb{Z}_{p^n} must be of the form \mathbb{Z}_{p^k} with 1\leq k <n. Suppose that:
\mathbb{Z}_{p^n} \cong \mathbb{Z}_{p^{k_1}} \times \cdots \times \mathbb{Z}_{p^{k_r}}. Consider the subgroups of this direct product:
\langle p^{k_1-1} \rangle \times \{0\} \times \cdots \times \{0\} and
\{0\} \times \cdots \times \{0\} \times \langle p^{k_r-1} \rangle
Both have order p. This is because p \cdot p^{k_1-1} = p^{k_1} = 0 in \mathbb{Z}_{p^{k_1}} and p \cdot p^{k_r-1} = p^{k_r} = 0 in \mathbb{Z}_{p^{k_r}}. Thus we have distinct subgroups of \mathbb{Z}_{p^{k_1}} \times \cdots \times \mathbb{Z}_{p^{k_r}} which is impossible if \mathbb{Z}_{p^{k_1}} \times \cdots \times \mathbb{Z}_{p^{k_r}} is cyclic because subgroups of cyclic groups are unique for each order. So we have a contradiction, thus \mathbb{Z}_{p^n} \ncong \mathbb{Z}_{p^{k_1}} \times \cdots \times \mathbb{Z}_{p^{k_r}}

4). Give an example to show that the weak direct product is not a coproduct in the category of all groups.
Consider the weak direct product \mathbb{Z}_2 \times \mathbb{Z}_3 with the canonical injections i_1: \mathbb{Z}_2 \rightarrow \mathbb{Z}_2 \times \mathbb{Z}_3 and i_2: \mathbb{Z}_3 \rightarrow \mathbb{Z}_2 \times \mathbb{Z}_3. Next define the group homomorphisms, \phi_1: \mathbb{Z}_2 \rightarrow S_3 by 1\mapsto (12) and \phi_2: \mathbb{Z}_3 \rightarrow S_3 by 1\mapsto (123). Now then we have a family of groups and maps into the weak direct product along with maps into a different group. In order for the weak direct product to be a coproduct we would now need a map from the weak direct product into the different group such that the diagram commutes. So suppose there exists \phi : \mathbb{Z}_2 \times \mathbb{Z}_3 \rightarrow S_3 such that \phi \circ i_j = \phi_j. Then we have that (12), (123) \in Im(\phi ). But these elements generate all of S_3 so that S_3 \cong Im(\phi ). But Im(\phi ) = \mathbb{Z}_2 \times \mathbb{Z}_3 which is abelian, while S_3 is not abelian. Therefore such a map does not exist.

5) Let G, H be finite cyclic groups. Then G\times H is cyclic iff gcd(|G|,|H|) =1
First we need to prove another fact. The order of an element (g,h) \in G \times H is l = lcm(|g|,|h|) Note that (g,h)^l = (g^l,h^l) = (e_G,e_H). Suppose there is some m <l s. t. (g,h)^m = (e_G,e_H). Then g^m = e_G and h^m = e_H, so the order of g divides m and the order of h divides m. This gives m as a common multiple of the orders of g and of h, but l is the least common multiple so that l \leq m which is a contradiction. Thus |(g,h)| = l. Now we can do the proof.
(\Rightarrow ) Assume that G\times H is cyclic so that G\times H = \langle (g,h) \rangle for some g\in G, h\in H. Let gcd(|G|,|H|) = d. Then we have:
(g,h)^{\frac{|G||H|}{d}} = (g^{\frac{|G||H|}{d}},h^{\frac{|G||H|}{d}}) = ((g^{|G|})^{\frac{|H|}{d}},(h^{|H|})^{\frac{|G|}{d}}) = (e,e). This tells us that |(g,h)| divides \frac{|G||H|}{d} but |(g,h)| = |G \times H | = |G||H| which means that d = 1.
(\Leftarrow ) Assume now that gcd(|G|,|H|) = 1. We know that G = \langle g \rangle and H = \langle h \rangle for some g, h. By the fact above we have
| \langle (g,h) \rangle | = |(g,h)| = lcm(|g|,|h|) = lcm(|G|,|H|) = |G||H| = |G \times H| Therefore (g,h) generates the entire group so that G\times H is cyclic.

6) Every finitely generated abelian group G \neq \langle e\rangle in which every non identity element has prime order p is isomorphic to \mathbb{Z}_p \bigoplus \cdots  \bigoplus \mathbb{Z}_p (n summands) for some natural number n.
Suppose G is finitely generated abelian and let X = \{ x_1,...,x_n\} be a minimal generating set of G. Since G is abelian all subgroups are normal so that \langle x_i \rangle \triangleleft G. Also G = \langle \cup_{i=1}^n \langle x_i \rangle \rangle. We need only show that \langle x_j \rangle \cap \langle \cup_{i \neq j} \langle x_i \rangle \rangle = \{e\}. Let a\in \langle x_j \rangle \cap \langle \cup_{i \neq j} \langle x_i \rangle \rangle and suppose a\neq e. Then it must be the case that \langle a \rangle \leq \langle x_j \rangle, but every element has order p by assumption so that p = |a| = |x_j| which gives that \langle x_j \rangle = \langle a \rangle  \leq \langle \cup_{i\neq j} \langle x_j \rangle \rangle. This tells us that anything in \langle x_j \rangle is also in \langle \cup_{i\neq j} \langle x_j \rangle \rangle which gives that x_j is redundant so G can be generated without the element x_j. This contradicts the minimality of the generating set. Thus a = e. Now from theorem 8.6 we can say G \cong \langle x_1 \rangle \times \cdots \times \langle x_n \rangle But each \langle x_i \rangle is cyclic of order p and hence isomorphic to \mathbb{Z}_p. Cyclic groups are abelian so we can rewrite this as G \cong \mathbb{Z}_p \bigoplus \cdots = \bigoplus \mathbb{Z}_p (n summands)

9) If a group G is the (internal) direct product of its subgroups H, K then H \cong G/K and G/H \cong K.
Let G = H \times K and consider the canonical projection maps:
\pi_H: H \times K \rightarrow H and \pi_K: H \times K \rightarrow K by \pi_H(h,k) = h and \pi_K(h,k) =k. Then note that \pi_H(x,y) = e iff y = e so that Ker\pi_H = K \times \{e\}. Clearly also \pi_H is surjective. Then by the first isomorphism theorem we have:
(H\times K)/Ker\pi_H \cong H \Rightarrow (H\times K)/(K \times \{e\}) \cong H \Rightarrow (H\times K)/K \cong H \Rightarrow G/K \cong H The other isomorphism is shown in the exact same manner.

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