Algebra Solutions: Hungerford II.1

1)(a) If G is an abelian group and m\in \mathbb{Z} then mG = \{mg | g \in G\} is a subgroup of G.
(b) If G = G_1 \bigoplus \cdots \bigoplus G_n, then mG = mG_1 \bigoplus \cdots \bigoplus mG_n and G/mG \cong G_1/mG_1 \bigoplus \cdots \bigoplus G_n/mG_n
Proof-
a) Since we can write 0 = m0 \in mG we have that mG is nonempty. Let x,y \in mG which means we can write x = mg_1 and y = mg_2 for some g_1,g_2 \in G. Then we have x - y = mg_1-mg_2 = m(g_1-g_2) \in mG so that it is a subgroup by the subgroup test.

b) Assume that G = G_1 \bigoplus \cdots \bigoplus G_n so that we can write for some g, g = (g_1,...,g_n) \Rightarrow mg = (mg_1,...,mg_n) \Rightarrow mG = mG_1 \bigoplus \cdots \bigoplus mG_n To show the next part we define f:G \rightarrow G_1/mG_1 \bigoplus \cdots \bigoplus G_n/mG_n by f(g) = (g_1 + mG_1,..., g_n + mG_n). Then this function is onto and we have it is a homomorphism by property of addition of cosets. Thus f is an epimorphism. Now let g \in Kerf this gives that f(g) = (mG_1,...,mG_n) which tells us that g_i +mG_i = mG_i for each i. This means that g_i\in mG_i for all i so that g\in mG_1 \bigoplus \cdots \bigoplus mG_n = mG. So that Kerf = mG. Then using the first isomorphism theorem we have G/mG \cong G_1/mG_1 \bigoplus \cdots \bigoplus G_n/mG_n

4) A free abelian group is a free group iff it is cyclic.
Proof-
(\Rightarrow ) Let G be a free abelian group that is also a free group. We know that any free group on a set X with |X| \geq 2 is nonabelian so we must have that |X| = 1. Every element in a free group has infinite order and X = \{a\} \Rightarrow G = \langle a \rangle.
(\Leftarrow ) Suppose that G is a free abelian group which is cyclic. Then G =  \langle a \rangle for some a\in G. Then each element of G has the form a^n for some integer n. Define f: G \rightarrow \mathbb{Z} by a^n\rightarrow n is a group isomorphism so that G is free since \mathbb{Z} is free.

10)(a) Show that the additive group \mathbb{Q} is not finitely generated.
(b)Show that \mathbb{Q} is not free
Proof-
Assume instead that\mathbb{Q} is finitely generated by some set A = (\frac{a_1}{b_1},...,\frac{a_k}{b_k}). Denote the product b_1 \cdots b_k = n. Let x\in \langle A \rangle so that we can write x = c_1 \frac{a_1}{b_1} + ... + c_k \frac{a_k}{b_k} for some c_i \in \mathbb{Z}. Then we can write x=\frac{m}{n} for some m\in \mathbb{Z}. This tells us that x\in \langle \frac{1}{n} \rangle. Thus \langle A \rangle \leq \langle \frac{1}{n} \rangle, but obviously \mathbb{Q} must also contain \langle \frac{1}{n} \rangle so that \langle A \rangle  = \mathbb{Q} = \langle \frac{1}{n} \rangle . Let a\in \mathbb{Z} such that gcd(a,n) = 1. Then \frac{1}{a} \in \mathbb{Q} = \langle \frac{1}{n} \rangle \Rightarrow \frac{1}{a} = \frac{b}{n} for some integer b. Finally we get that ab = n which contradicts gcd(a,n) = 1. Thus the rational numbers are not finitely generated.
b) Let x=\frac{a}{b}, y = \frac {c}{d} be any two rationals. If mx + ny = 0, then we have m\frac{a}{b} + n\frac {c}{d} = 0 and we can have m =bc and n = -ad. This shows x and y are not linearly independent. Thus if, \mathbb{Q} was a free group, it would have to be of rank 1, which would make it cyclic, but from part a) we know that it is not cyclic (because it is not finitely generated). Thus the rationals are not a free group.

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