Hw8 Problem 12

(a) Suppose $\mid G \mid = 2n, n \in \mathbb{Z}.$ Show that $G$ contains an element of order $2$. (i.e., $\exists a \in G$ such that $a*a=e$.)

(b) Use Lagrange's Theorem to show that if $G$ is abelian and $\mid G \mid = 2n$ where $n$ is odd, then $G$ contains exactly one element of order $2$.

Solution
(a)
Suppose $\mid G \mid = 2n, n \in \mathbb{Z}^+.$
Now, Suppose that $G$ does not contain an element of order $2$.
Then $\nexists a \in G$ such that $a*a=e$ with $a \not = e$.
That is, $\nexists a \in G$ such that $a=a^{-1}$ with $a \not = e$.
Observe that by properties of groups, $e\in G$, and $\forall a \in G, a^{-1} \in G.$
Hence, $G$ may be written as $\{ e, a_1, a_1^{-1}, a_2, a_2^{-1}, ..., a_m, a_m^{-1}\}$ with $m \in \mathbb{Z}^+$.
Now, since we have already shown that $a_i\not =a_i^{-1} \ \forall a_i \in G, \mid G\mid = 1 + 2m$ with $m \in \mathbb{Z}^+$. But, we assumed the order of $G$ was even, so we have a contradiction.
Therefore, $G$ contains an element of order $2$.

(b)
Suppose $G$ is an abelian group with $\mid G\mid =2n$ with $n$ an odd integer.
Then $\not \exists k \in \mathbb{Z}$ such that $\mid G \mid = 4k$.
Hence, $\mid G\mid$ is not divisible by $4$.
Thus, by Lagrange's Theorem, there cannot be a subgroup of $G$ with order $4$.
Now, by part (a), $\exists a \in G$ such that $\mid \langle a \rangle \mid = 2$ and specifically, $a= a^{-1}$.
Suppose that there is another such element, $b\in G$, with $\mid \langle b \rangle \mid = 2$.
Then, by the same reasoning used in part (a), $b = b^{-1}$.
Now, consider the set $S=\{e, a, b, ab\}$.
Since G is abelian, any subgroup of G is abelian.
Thus S is abelian, and this may be used to show the following:

$\forall x \in S, x*e=e*x = x$, by definition of identity, so S is closed under multiplication by $e$.
$b*a=ba=ab \in S$
$a*b = ab \in S$
$b*b=bb=e \in S$
$b*ab=bab=abb=ae=a\in S$
$ab*b = abb = ae = a \in S$
$a*ab = aab = eb = b \in S$
$ab*a=aba=aab=eb=b \in S$
$ab*ab=abab=aabb=ee=e \in S$

Thus, $S$ is closed under the operation $*$.
Observe that by the construction of $S$, the identity element from $G$ is in $S$.
Further observe that in the calculations above, each element was shown to be its own inverse.
Hence each element of $S$ has an inverse in $S$.
Thus, since all conditions hold, $S \leq G$.
However, by observation, $\mid S \mid = 4$, which contradicts the previous statement that no such subgroup exists.
Therefore, $a$ is the unique element in $G$ with order $2$.
$\hspace{250pt} \clubsuit$