Exercise 2.1.2 (Subsets which are not subgroups)

Proof.

(a)

Suppose that $n\ge 3$. Then the two $2$-cycles $(12)$ and $(23)$ are in $S_n$, but $$(12)(23)=(123)$$

is not a $2$-cycle.

The set of $2$-cycles in $S_n$ is not a subgroup of $S_n$ for $n\ge 3$ (even if we add the identity element in this set).

(b)

Consider the two reflections $s$ and $\mathit{rs}$ in $D_{2n}$ (where $r^n=s^2=e$). Then $$(\mathit{rs})s=r$$

is of order $n\ge 3$, so is not a reflection.

The set of reflections in $D_{2n}$ is not a subgroup of $D_{2n}$ (even if we add the identity element in this set).

(c)

Put $H=\{x\in G\mid |x|=n\}\cup \{1\}$.

Since $n>1$ is composite, $n=\mathit{ab}$ for some integers $a,b$ such that $a>1$, $b>1$.

By hypothesis, $G$ contains an element $x$ of order $n$, so $x\in H$. But $x^a$ has order $b$: indeed, for all integers $k$,

$${(x^a)}^k=1⟺x^{\mathit{ak}}=1⟺n\mid \mathit{ak}⟺\mathit{ab}\mid \mathit{ak}⟺b\mid k.$$

From $1<b<n$, we infer that $x\ne 1$ and $x$ has not order $n$, so $x^a\notin H$. Therefore $H$ is not a subgroup of $G$.

(d)

Let $H$ be the set of odd integers in $\mathbb{Z}$ together with $0$. Then $7\in H$ and $3\in H$, but $7-3=4\notin H$. Therefore $H$ is not a subgroup of $\mathbb{Z}$.

(e)

Let $H$ be the set of real numbers whose square is a rational number: $$H=\{x\in ℝ\mid x^2\in \mathbb{Q}\}.$$

Then $1\in H$ and $\sqrt{2}\in H$ (because $1^2=1\in \mathbb{Q}$ and ${\sqrt{2}}^2=2\in Q$. But

$$a={(1+\sqrt{2})}^2=3+2\sqrt{2},$$

and $a=3+2\sqrt{2}\notin \mathbb{Q}$ (otherwise $\sqrt{2}=(a-3)∕2$ would be rational).

Since $1\in H$ and $\sqrt{2}\in H$, but $1+\sqrt{2}\notin H$, $H$ is not a subgroup of $(ℝ,+)$.

(But $\{x\in {ℝ}^{\text{×}}\mid x^2\in \mathbb{Q}\}$ is a subgroup of $({ℝ}^{\text{×}},\times )$ see Exercise 1 (e)).