**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.1 Exercise 2.1.1**

For each of the following, show that the specified subset is a subgroup of the given group.

(1) The set $A$ of complex numbers of the form $a + ai$ under addition.

(2) The set $A$ of complex numbers with absolute value 1 under multiplication.

(3) For a fixed $n \in \mathbb{Z}^+$, the set $A$ of rational numbers whose denominators divide $n$, under addition.

(4) For a fixed $n \in \mathbb{Z}^+$, the set $A$ of rational numbers whose denominators are relatively prime to $n$, under addition.

(5) The set $A$ of nonzero real numbers whose square is a rational number, under multiplication.

Solution:

(1) First, $A$ is nonempty since $0=0 + 0i \in A$. Now if $a+ai$, $b+bi \in A$, we have $$(a+ai) - (b+bi) = (a-b) + (a-b)i \in A.$$ By the subgroup criterion, $A$ is a subgroup of $(\mathbb{C}, +)$.

(2) Note that $A$ is not empty since $|1+0i| = |1| = 1$. Let $a+bi$, $c+di \in A$; then $$a^2+b^2 = c^2+d^2 = 1.$$ Then \begin{align*}(a+bi)(c+di)^{-1} =&\ \frac{a+bi}{c+di}\\ =&\ \frac{(a+bi)(c-di)}{c^2+d^2} \\=&\ (ac+bd) + (bc-ad)i,\end{align*} which has absolute value \begin{align*}&\ (ac+bd)^2 + (bc-ad)^2 \\=&\ a^2c^2 + 2abcd + b^2d^2 + b^2c^2 - 2abcd + a^2d^2\\ =&\ (a^2+b^2)(c^2+d^2) = 1.\end{align*} By the subgroup criterion, then, $A$ is a subgroup of $(\mathbb{C}, \cdot)$.

(3) Note that $A$ is not empty since $0 = 0/1 \in A$. Let $a/q$, $b/p \in A$; then $q|n$ and $p|n$. Let $d = \mathsf{gcd}(q,p)$, and write $q = kd$ and $p = \ell d$. Now $$\frac{a}{q} - \frac{b}{p} = \frac{ap-bq}{qp} = \frac{a \ell d - bkd}{kd \ell d} = \frac{a \ell - bk}{k \ell d} = \frac{a \ell - bk}{\mathsf{lcm}(q,p)}.$$ Clearly $\mathsf{lcm}(q,p)$ divides $n$, so by the subgroup criterion, $A$ is a subgroup.

(4) First we prove a lemma.

**Lemma.** If $\mathsf{gcd}(a,c) = \mathsf{gcd}(b,c) = 1$, then $\mathsf{gcd}(ab,c) = 1$.

Proof: We have $qx + ny = 1$ and $pz + nw = 1$ for some integers $x,y,z,w$. Then $$1 = (qx+ny)(pz+nw) = qp(xz) + n(qxw + ypz + nyw), $$so that by a lemma to a previous exercise, $\mathsf{gcd}(ab,c) = 1$. $\blacksquare$

Now $A$ is not empty since $0 = 0/1 \in A$. Let $a/q$, $b/p \in A$; then $$\mathsf{gcd}(q,n) = \mathsf{gcd}(p,n) = 1.$$ By the lemma, $\mathsf{gcd}(qp,n) = 1$, so that $$\frac{a}{q} - \frac{b}{p} = \frac{ap-bq}{qp} \in A.$$ By the subgroup criterion, $A$ is a subgroup.

(5) Note that $A$ is not empty since $1^2 \in \mathbb{Q}$. Now if $x,y \in A$, we have that $x^2 = q$ and $y^2 = p$ for some rational numbers $p$ and $q$. Then $$(x/y)^2 = x^2/y^2 = q/p \in \mathbb{Q};$$ by the subgroup criterion, $A$ is a subgroup.