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Demonstrate that a given subset is not a subgroup


Show that in each of the following examples, the specified subset is not a subgroup.

(1) The set of 2-cycles in $S_n$ for $n \geq 3$.
(2) The set of reflections in $D_{2n}$ for $n \geq 3$.
(3) For $n$ a composite integer greater than $1$ and $G$ a group containing an element of order $n$, the set $\{ x \in G \ |\ |x| = n \} \cup \{ 1 \}$.
(4) The set of all odd integers with 0.
(5) The set of all real numbers whose square is rational, under addition.


Solution:

(1) Note that $(1\ 2)(2\ 3) = (1\ 2\ 3)$, so this set is not closed under multiplication.

(2) Note that since $n \geq 3$, we have $sr^{-1} \cdot sr^2 = r$. Thus this set is not closed under multiplication.

(3) Let $x \in G$ have order $n$, where $n = ab$. Then $x^a$ has order $b$, and so is not in $G$; this is a contradiction.

(4) Note that $1+1 = 2$, so this set is not closed under addition.

(5) Note that $(\sqrt{2} + \sqrt{3})^2 = 5 + 2 \sqrt{6}$, which is not rational. So this set is not closed under addition.

Linearity

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