GATE CSE 1994 | Question: 1.10

Question

Some group $(G, o)$ is known to be abelian. Then, which one of the following is true for $G$?

• A. $g=g^{-1} \text{ for every } g \in G$

• B. $g=g^2 \text{ for every }g  \in G$

• C. $(goh)^2 = g^2oh^2 \text{ for every } g, h \in G$

• D. $G$ is of finite order

Comments

i think option 1 and 3 both are correct because 
 (a*b)^-1 = b^-1*a^{-1
so, for x-1 = x}

L.H.S.=

(a*b)^-1=a*b

R.H.S.=

b^-1 * a^-1 = b*a
which satisfies commutative property means option 1 is also correct
 

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@suryansh rajput 

( {1, -1, i, -i } , $\times$ ) is abelian group with identity element 1. inverses of 1, -1, i, -i are 1, -1, -i, i respectively. so the first option is not correct.

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opt (B) also tells
g o g = g
=>$g^-1$ o g o g=$g^-1$ o g
=>($g^-1$ o g) o g=$g^-1$ o g
=>e o g = e
Again, we know e is identity ele.
So e o g =g ------→ g = e
So opt(b)  tells every element is identity ele in abelian grp, which is wrong.
Because there exists atmost 1 identity ele in an algebraic structure.

Answer 1

Associativity property of Group.

• For all $a, b$ and $c$ in $G$, the equation $(a o b) o c = a o (b o c)$ holds.

For an Abelian group, commutative property also holds. 

• For all $a, b$ in $G$, the equation $a o b = b o a$

From option C, using these two properties,

$(goh)^2  = (goh) o (goh) \\= (h o g) o (g o h)  \\= ((h o g) o g ) o h \\= (h o (g o g)) o h \\= (h o g^2) o h \\= (g^2 o h) o h \\= g^2 o (h o h) \\= g^2 o h^2$

So, C is correct.   

Integer addition $(Z, +)$ is an Abelian group. 

Inverse of $1$ is $-1$ and not $1.$ So, A is false. 

$1^2 = 1 + 1 = 2 \neq 1$. So B also false. 

The order of a group is the number of elements in it. An integer is an infinite set, so D is also false. 
References:

•  http://math.stackexchange.com/questions/40996/prove-that-if-abi-aibi-forall-a-b-in-g-for-three-consecutive-integers/41004#41004
•  http://math.stackexchange.com/questions/423745/a-group-g-is-abelian-iff-abn-an-bn-for-all-a-b-in-g-and-n-in-bb?lq=1

Comments

how is option c satisfied for (Z, +)?

eg: ($2+3)^{2} \neq 2^{2} + 3^{2}$

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@akshay7797

The power doesn't mean product. It means the same operation is done multiple times.

In case of $(Z, +)$,
$LHS = (g + h) + (g + h) = 2(g + h)$
$RHS = (g + g) + (h + h) = 2(g + h)$

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How we come to know that its a set of integers

Answer 2

Some common Abelian groups:-

• (Z, +)
• (R, +)
• (Ev, +) //Set of Even numbers on addition
• (M, +) //Matrix addition

For addition, identity element, e = 0.

Inverse would be anything that results in 0 after addition.

 

Let's take the set of even numbers.

 

Is 2 = -2 ? No. So, Option A

Is 3 = 6? No. So, Option B //Note that here, $3^2=3+3=6$

Is set of even numbers finite? No. So, Option D

Comments

great !

Answer 3

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