Problem 1. Let $G$ be a finite group. Show that the number of elements in $G$ of order greater than 2 must be even. Conclude that any group of even order must contain an element of order 2.

Solution. This is really just a simple counting argument. Let $S$ be the set of all elements of $G$ with order greater than two. If $S=\emptyset$ then $|S|=0$ and we’re finished. So assume $S \neq \emptyset$. Since $|x|=\left|x^{-1}\right|$ for any $x \in G$, we see that $x \in S$ if and only if $x^{-1} \in S$. Moreover, if $x \in S$ then $x^2 \neq e$ so that $x \neq x^{-1}$. It follows that $S$ can be written as the disjoint union of two element sets of the form $\left{x, x^{-1}\right}$, and hence that $|S|$ is even.

Suppose that $|G|$ is even. Since $e$ has order $1, e \notin S$. It follows that $G \backslash S \neq \emptyset$. So $0<|G \backslash S|=|G|-|S|$. Since $|G|$ and $|S|$ are both even, it follows that $|G \backslash S|$ is a nonzero even integer, i.e. is at least 2. Thus, there is an $x \in G \backslash S, x \neq e$. Since $S$ consists of all elements in $G$ of order greater than 2 , it must be the case that $|x|=2$.

Problem 2. A group $G$ is called divisible if given any $x \in G$ and any $n \in \mathbb{Z}^{+}$there exists a $y \in G$ so that $y^n=x$.
(a) Show that $\mathbb{Q}$ (with the operation of addition) is divisible.
(b) Show that a cyclic group is never divisible.

Solution. (a) Let $x \in \mathbb{Q}$ and $n \in \mathbb{Z}^{+}$. Then $y=x / n \in \mathbb{Q}$ and
$$n y=n \frac{x}{n}=\underbrace{\frac{x}{n}+\frac{x}{n}+\cdots+\frac{x}{n}}_{n \text { terms }}=x$$
which expresses the divisibility property in additive form.
(b) Let $G=\langle g\rangle$ be a cyclic group. Suppose first that $G$ is infinite. Take $x=g$ and $n=2$. For any $y \in G$ there is an $m \in Z$ so that $y=g^m$ so that $y^2=g^{2 m}$. Thus, if $y^2=g$ then $2 m=1$, which is impossible. Therefore, $G$ is not divisible.

If $G$ is finite, choose any $x \in G, x \neq e$, and let $n$ be the LCM of all of the orders of elements in $G$. Then for any $y \in G$ we have $y^n=e \neq x$, so that $G$ is not divisible. Note that in the finite case we never had to use the fact that $G$ was cyclic!

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