GO Classes Test Series 2024 | Discrete Mathematics | Test 3 | Question: 21

Answer 1

“If for all $x \in \text{G} , x^{2} = 1,$ then $\text{G}$ is commutative. Here, $1$ is the identity element of $\text{G}.$” : True

$\textbf{Proof 1 :}$
Let $a,b$ be two elements in group $\text{G}.$ Consider the element $ab.$
Now, $(ab)^{2} = 1$
$ab.ab = 1$
Now, multiply both sides on the right with $b,$
$aba=b$
Now, multiply both sides on the right with $a,$
$ab=ba$
Hence, $\text{G}$ is abelian.

$\textbf{Proof 2:}$
Let $a,b$ be two elements in group $\text{G}.$
It is given that $(ab)^{2}=1 ;$ So, since $(ab)(ab)=1,$ So, $ab$ is inverse of $ab.$
Now consider $(ab)(ba).$
$(ab)(ba)=a(bb)a=aa=1 ;$ So, $(ab)(ba)=1 ;$ which means that $ba$ is inverse of $ab.$
But every element in a group has unique inverse, So, $ab=ba.$
Hence, $\text{G}$ is abelian.

Reference: https://math.stackexchange.com/questions/238171/prove-that-if-g2-e-for-all-g-in-g-then-g-is-abelian

Comments

It is similar to “If every element of the group G is its own inverse then G is abelian but the converse is not true.”