## 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.