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Let (G,*) be a group such that O(G) = 8, where O(G) denotes the order of the group. Which of the following is True ?

  1. There exist no element a in G whose order is 6.
  2. There exist an element a in G whose order is 4.
  3. There exist more then one element in G whose order is 1
  4. None of these
in Set Theory & Algebra
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A?
0
Please explain the reason for every statment.

Answer is correct though :)

2 Answers

6 votes
 
Best answer

Let $G$ be a finite group of order $k$

then every element of $G$ will have order $m$ such that $k \ \text{mod } m = 0$

 

A is true because $8 \ \text{mod } 6 = 2$ 

B is false because $8 \ \text{mod } 4 = 0$  but it is not necessary that there must exist an element with order 4.

for example Elementary abelian group:E8 is of order 8 but does not contain any element of order 4

is false because there must exist only 1 identity element in a group


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For B the correct option should be that it may contain an element..... Right sir ?
1
yes may or may not be

sir mat bolo bhai :p
0

Let G be a finite group of order k

then every element of G will have order m such that k mod m=0

 @

which theorem is this?plz explain!

0

@

Thanks..got it!

0 votes
Order of an element in group always divides order of group.

8%6 !=0

So there does not exist any element with order 6.

A is true

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