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

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

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

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

  4. $G$ is of finite order

in Set Theory & Algebra by Veteran | 3.2k views
D is false. (Z,+) is an infinite abelian group. Don't know about other options.
how option b is incorrect ?
try for (z,+mod) you got why its wrong

2 Answers

+50 votes
Best answer

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. 

Order of a group is the number of elements in it. Integer is an infinite set, so D is also false. 
ref @

by Veteran

Another way of proving option C

For Groups we know cancellation property holds, i.e. 

a.b=a.c ⇒ b=c (why this is true ?, because we have inverse element for each group. see this)

Having known this,

(goh)2=(goh)o(goh) = gohogoh (im removing bracket because groups are associative ).

Now LHS becomes gohogoh. and RHS is g2oh2. We want to proof gohogoh = g2oh2.

 gohogoh = g2oh2

ohogoh =goh2 (cancel leftmost g from each sides, then it becomes hogoh = goh2)

Similarly, cancel rightmost h from both sides. then it becomes hog = goh.

which is true for abelian groups, since abelian groups are commutative.


Very Nice Explanation !

There is a theorem:  $g=g^{−1}$ for every g∈G $\implies$ G is abelian.

But it is not two way i.e. converse is not true.

well explained.Thanks :-)


(goh)2 = (goh)o(goh) how??

Please explain


similar to (3*4)= (3*4)*(3*4).


@Arjun Sir..How (gog)=g?

lovely explanation sir

Proving from RHS,

g2oh2 = gog o hoh = go (goh) o h= go hog oh [As abelian so goh=hog]

(goh) o (goh) = (goh)2

In case of (M,+) where M is set of all Matrices.

option C will not be true for any two matrices right? Consider two matrices

A= 1   2      B= 1  1  Here (A+B)^2 is not equal to A^2+B^2 right?

      3   4           0  1

P.S: We know that (M,+) is an Abelian group.

Anyone please clarify.
how is option c satisfied for (Z, +)?

eg: ($2+3)^{2} \neq 2^{2} + 3^{2}$
+1 vote

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

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