## Q* is residually finite

Posted: May 13, 2010 in Elementary Algebra; Problems & Solutions, Groups and Fields
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Definition. A group $G$ is called residually finite if for every $1 \neq g \in G$ there exists a finite group $H$ and a group homomorphism $\varphi : G \longrightarrow H$ such that $\varphi(g) \neq 1.$

Example 1. A subgroup of a residually finite group is residually finite.

Example 2. Every finite group is residually finite.

Proof. For a given $g \in G$ we may choose $H=G$ and $\varphi = \text{id}_G.$

Example 2. $\mathbb{Z}$ is residually finite.

Proof. For a given integer $n$ we may choose $H = \mathbb{Z}/p \mathbb{Z},$ where $p$ is any prime number not dividing $n.$ Define $\varphi : \mathbb{Z} \longrightarrow H$ by $\varphi(m)=m+p\mathbb{Z},$ for all $m \in \mathbb{Z}.$ Clearly $\varphi(n) = n + p \mathbb{Z} \neq 0$ because $p \nmid n.$

Example 3. A direct sum or product of residually finite groups is residually finite.

Proof. Let $\{G_i \}_{i \in I}$ be a family of residually finite groups and put $G= \oplus_{i \in I} G_i.$ Let $1 \neq g = (g_i) \in G.$ So $g_k \neq 1$ for some $k \in I.$ Also, there exists a finite group $H_k$ and a group homomorphism $\varphi : G_k \longrightarrow H_k$ such that $\varphi_k(g_k) \neq 1.$ Now, for any $k \neq i \in I$ let $H_i = \{1\}$ and define $\varphi_i : G_i \longrightarrow H_i$ by $\varphi_i(x_i)=1,$ for all $x_i \in G_i.$ Finally define $\varphi : G \longrightarrow \bigoplus_{i \in I} H_i$ by $\varphi (x_i) = (\varphi_i(x_i)),$ for all $x = (x_i) \in G.$ Clearly $\varphi(g) \neq 1$ because $\varphi_k(g_k) \neq 1.$

Example 4. $\mathbb{Q}^{\times},$ the multiplicative group of $\mathbb{Q},$ is residually finite.

Proof. Let $G=\{1,-1\} \oplus \mathbb{Z} \oplus \mathbb{Z} \oplus ...,$ where $\{1,-1\}$ is considered as a subgroup of $\mathbb{Q}^{\times}.$ Let $p_j$ be the j-th prime number. Define $f : \mathbb{Q}^{\times} \longrightarrow G$ by

$f(x)=(\text{sgn}(x),n_1,n_2, \cdots , n_k, 0, 0, \cdots),$

where $x=\pm p_1^{n_1}p_2^{n_2} \cdots p_k^{n_k}.$ Clearly $f$ is a group isomorphism. The result now follows from Example 2 and 3.