Question

If G is an abelian group of order 2n, withn odd, prove that G contains exactly one element of order 2.

Short answer

We proved that G contains one element of order 2

Step-by-step solution

To find another element of the group of order 2 

Let G be an abelian group of even order 2n with n odd.

To prove that Gcontains exactly one element of order 2.

Since the order of Gis even, it follows from Exercise 31 of the present section that G contains one element of order 2, say $X\in G$$X\in G$.

Now, we have to show that if there is another element $Y\in G$of order 2, then .$x=y$

If possible, let role="math" localid="1652258381725" $\exists y\in G\backslash \left\{x,e\right\}$ of order 2, where is an identity element.

Then

${\left(xy\right)}^2=xyxy$

$=x^2{y}^2$ ...(G is abelian)

$=$e ....(both x and y have order 2)

(xy)²$=$e

To prove G  contains exactly one element of order 2

Next, consider subset $K=\left\{e,x,y,xy\right\}$ andKis a subgroup of .

Here, the order of K is 4.

Thus, by Lagrange’s Theorem, 4 divides the order of G .

i.e., 4|2n …. (o(G)=2n )

2|n

This is a contradiction to the fact that n is odd.

Hence, we have x=y .

Hence, G contains one element of order 2.