GATE CSE
First time here? Checkout the FAQ!
x
+2 votes
113 views

A group G in which (ab)2 = a2b2 for all a,b in G is neccessarily

A. finite
B. cyclic
C. of order two
D. Abelian

please prove it
ands: D   

asked in Set Theory & Algebra by Veteran (14.6k points)   | 113 views

2 Answers

+2 votes
Best answer

Abelian can be proved as follows:

answered by Active (1.7k points)  
selected by
what is identity element and inverse?
@Vijay. Why do you need identity and inverse? They have already given that its a group. So, they exist.

#Sushant Gokhale ? do you have any other idea of proving it to follow commutative property. If yes..please share

@debashish. We need to assume the operator here since they havent given the operator. So, your answer as well as above answer is correct.

Other approach:

(ab)2 = a2b2

$\therefore$ abab = aabb

 

Now, for group cancellation law holds.

$\therefore$ bab = abb

$\therefore$ ba = ab

 

Thus, proved.

"They have already given that its a group" yes, you're right then will surely exist
+3 votes
$$\begin{align*} & {\color{blue}{(S \;, \;\#)}}\;\; \text{ : is a Abelian Group if it satisfies all the following properties } \\ \\ &\begin{cases} & 1.\text{ Closure} \\ & 2.\text{ Associative} \\ & 3.\text{ Identity element exits} \\ & 4.\text{ Inverse for all elements} \in S \\ & 5.\text{ Commutative} \\ \end{cases} \\ \\ \hline \\ &\text{ A } {\color{red}{\bf \text{group}}} \text{ satisfies properties } \;\; {\color{blue}{1,2,3,4}} \\ &\text{ When a group } \;\; {\color{blue}{(S \;, \;\#)}}\;\; \text{satisfies commutative property} \\ &\Rightarrow {\color{blue}{(S \;, \;\#)}}\;\; \text{is } {\color{red}{\bf \text{Abelian}}} \text{ also.} \\ \\ \hline \\ \end{align*}$$

$\begin{align*} &\Rightarrow \text{LHS} = \left ( a \; \# \; b \right )^2 \\ &\Rightarrow \text{LHS} = \left ( a \; \# \; b \right ) \# \left ( a \; \# \; b \right ) \\ &\Rightarrow \text{LHS} = a \; \# \; \left (\bf b \; \# \; a \right ) \; \# \; b \\\\ \hline \\ &\Rightarrow \text{RHS} = a^2 \; \# \; b^2 \\ &\Rightarrow \text{RHS} = \left ( a \; \# \; a \right ) \# \left ( b \; \# \; b \right ) \\ &\Rightarrow \text{RHS} = a \; \# \; \left (\bf a \; \# \; b \right ) \; \# \; b \\\\ \hline \\ &\text{Given LHS = RHS} \\ &\Rightarrow \left ( b \; \# \; a \right ) = \left ( a \; \# \; b \right ) \\ &\Rightarrow \text{Given group follows commutative property }\\ &\Rightarrow \text{Given group is} {\color{red}{\bf \text{Abelian}}} & \\ \end{align*}$
answered by Veteran (43.9k points)  


Top Users Mar 2017
  1. rude

    4768 Points

  2. sh!va

    3054 Points

  3. Rahul Jain25

    2920 Points

  4. Kapil

    2728 Points

  5. Debashish Deka

    2602 Points

  6. 2018

    1574 Points

  7. Vignesh Sekar

    1422 Points

  8. Akriti sood

    1378 Points

  9. Bikram

    1342 Points

  10. Sanjay Sharma

    1126 Points

Monthly Topper: Rs. 500 gift card

21,517 questions
26,845 answers
61,157 comments
23,181 users