The number of elements of order 5 in a non-abelian group of order 10 isa)1b)5c)4d)8Correct answer is option 'C'. Can you explain this answer?

Question

Answer

Any group of order 5 is cyclic .
so number of element will be phi 5 = 4

Answer

Let the group be GG. Observe that at least one elment of the group will have order 55 as:

No element can have order 1010 otherwise the group will become cyclic and hence abelian.

Not all non-identity elements will be of order 22 otherwise (ab)2=e⟹abab=e⟹ba=a−1b−1=ab(ab)2=e⟹abab=e⟹ba=a−1b−1=ab. So the group is abelian.

The group can have only elements of order 11, 22, 55, 1010. Since we ruled out possibility of all non-identity elements being of order 22 and none of them are of order 1010, there must be at least one of order 55.

Let o(a)=5o(a)=5. Then the subgroup generated by it will have 55 elements, one identity and 44 of order 55 (why?).

Suppose b∉b∉ and o(b)=5o(b)=5, then ∩={e}∩={e}. Let H=H= and K=K= then o(HK)=o(H)o(K)o(H∩K)=25>o(G)o(HK)=o(H)o(K)o(H∩K)=25>o(G). This is absurd and not possible.

So all the elements of order 55 are in . So there are exactly 44 of them.

Answer

Explanation:

In order to determine the number of elements of order 5 in a non-abelian group of order 10, we can use the concept of conjugacy classes and the class equation.

Conjugacy Classes:
In a group, the conjugacy class of an element 'a' consists of all elements 'x' such that 'xax⁻¹ = a'. In other words, the conjugacy class of 'a' is the set of all elements that are conjugates of 'a'.

Class Equation:
The class equation states that the order of a group is equal to the sum of the sizes of its conjugacy classes. Mathematically, if G is a group and C₁, C₂, ..., Cₙ are its conjugacy classes, then |G| = |C₁| + |C₂| + ... + |Cₙ|.

Applying the Class Equation:
Let's consider a non-abelian group of order 10. Since the group is non-abelian, it must have at least one non-trivial conjugacy class (i.e., a conjugacy class other than {e}, where e is the identity element).

Let's assume that there are k elements in the non-trivial conjugacy class. Since each element's conjugacy class has a size that divides the order of the group, the size of the non-trivial conjugacy class must be either 2, 5, or 10.

If the size of the non-trivial conjugacy class is 2, then the class equation becomes 10 = 1 + 2k, which implies k = 4. However, in this case, the group would be abelian, which contradicts the given condition.

If the size of the non-trivial conjugacy class is 10, then the class equation becomes 10 = 1 + 10k, which is not possible since k cannot be a fraction.

Therefore, the size of the non-trivial conjugacy class must be 5. This means that there are 5 elements in the group that are conjugates of each other.

Elements of Order 5:
Now, we know that the size of the non-trivial conjugacy class is 5. The order of an element in a group is equal to the size of its conjugacy class. Therefore, there are 5 elements in the group that have order 5.

Conclusion:
The number of elements of order 5 in a non-abelian group of order 10 is 5. Therefore, the correct answer is option C.

Answer

G=D5 then no. of element of order 5 = phi(5) = 4

Answer

Option c is right

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