GATE CSE 2022 | Question: 17

Question

Which of the following statements is/are $\text{TRUE}$ for a group $\textit{G}?$

• A. If for all $x,y \in \textit{G}, \; (xy)^{2} = x^{2} y^{2},$ then $\textit{G}$ is commutative.
• B. If for all $x \in \textit{G}, \; x^{2} = 1,$ then $\textit{G}$ is commutative. Here, $1$ is the identity element of $\textit{G}.$
• C. If the order of $\textit{G}$ is $2,$ then $\textit{G}$ is commutative.
• D. If $\textit{G}$ is commutative, then a subgroup of $\textit{G}$ need not be commutative.

Answer 1

Answer : A,B,C

There are many equivalent definitions (if and only if statements) for Abelian Groups.

In the GATE exam so far, 3-4 definitions have been asked, but some more variations can be created.

In the following video, I have covered 7-8 definitions of Abelian Groups, with proofs. The following video also contains $\color{red}{\text{detailed video solution}}$  of this GATE 2022 Commutative Group question.

Alternative Definitions/Variations of Abelian Groups

In Abstract Algebra, “Abelian” word is used as a synonym of “commutative”. So, Commutative Group is same as Abelian Group. (Abelian groups are named after early 19th century mathematician Niels Henrik Abel.)

Option D :

“If group $G$ is commutative, then a subgroup of $G$ need not be commutative” : False.

Theorem : $\color{blue}{\text{ Every subgroup of an abelian group has to be abelian. }}$
Proof :- Let $G$ be an abelian group, and suppose that $H ≤ G.$ 
Now we have to check that,  for any $a, b \in H,$ Do we have $ab = ba$ ??

Since we have : $a, b \in H \subseteq G$, So $a,b$ are elements of $G$ and since $G$ is abelian, So, $ab = ba.$
Therefore, $H$ is abelian as well.

https://math.stackexchange.com/questions/3198960/does-every-subgroup-of-an-abelian-group-have-to-be-abelian

Option C :

“If the order of group $G$ is 2, then $G$ is commutative” : True.

There is only one abstract group of order 2 possible. Or we can say that ALL the groups of order 2 are isomorphic(similar) to each other.

There is only one abstract group of order 2 possible and It can be easily seen that group of order 2 is abelian. 
We can similarly see that every group of order less or equal to five is abelian.

https://math.stackexchange.com/questions/1284709/prove-every-group-of-order-less-or-equal-to-five-is-abelian

Option B :

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

Proof 1 :
Let $a,b$ be two elements in group $G.$ 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, $G$ is abelian. 

Proof 2 :
Let $a,b$ be two elements in group $G.$
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, $G$ is abelian.

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

Option A :

$”$ If for all $x,y \in G$ , $(xy)^2 = x^2y^2$ , then $G$ is commutative $”$. : True.

Since $(ab)(ab) = (aa)(bb)$
We know that in a group, left and right cancelation is allowed; So, 
Cancel left most $’a’$ on both sides ;  Cancel right most $’b’$ on both sides; 
We get $ba = ab.$
 

Comments

Excellent Explanation Sir. 

For Option C: Every group of order $<=5$ is abelian. Infact, the least non-abelian group order $=6$

Answer 2

Answer : Option A,B and C are true.

---

Option A : If for all x,y ∈ G, $(xy)^2 = x^2.y^2 $, then G is commutative

$(xy)^2 = x^2.y^2 $

$xy . xy = xx.yy $

$apply\; both\; sides\; x^{-1} $

$x^{-1}.xy . xy = x^{-1}xx.yy $

$y . xy = x.yy $

$apply\; both\; sides\; y^{-1} $

$x.yy^{-1} . xy = x.yy^{-1} $

$y . x = x.y $

therefore G is comutative.

 

Option B : If for all x ∈ G, $x^2 = 1 $, then G is commutative. Here 1 is the identity element of G

$(x)^2 = Identity \;element $ that means x is inverse to itself.

$(x.y)^{-1} =  y^{-1}. x^{-1} $

$(x.y)^{-1} =  y. x $

$(x.y) =  y. x $

therefore G is commutative

 

Option C : If the order of G is 2, then G is commutative.

Theorem :- If Group order is prime number, then it is cyclic group. And we know that, every Cyclic group is abelian group.

Alternatively, We can think like, Given that Order of Group G is 2, one element is Identity element, So another element x is inverse to itself. By Option B, G is commutative.

 

Option D : If  G is commutative, then a subgroup of  G need not be commutative. – False

correct statement : If  G is commutative, then a subgroup of  G should be commutative.

why ? let (G,.) is a commutative group with elements {a,b,c,d} 

we know that a.b = b.a = c. If sub group H contains both a and b, then how can we get a.b = c, b.a = d  and c ≠ d ?