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+2 votes
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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 (14.7k points)   | 118 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 (48k points)  


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