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If p is a prime number and G is non-abelian group of order p3, then the centre of G has
- a)Exactly p - 1 elements
- b)Exactly p elements
- c)Exactly p3 element
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
If p is a prime number and G is non-abelian group of order p3, then th...
Centre of G is 
Since 0(G) = p3, p is a prime number
But Z is a subgroup of G => 0 ( Z ) / 0 ( G ) i.e. 0 ( Z )
divides 0(G) ⇒ 0 (Z) / p3
⇒ 0 ( Z ) = p or p2 or p3.
If 0 (Z) = p3 = 0 (G) ⇒ Z = G => is abelian co ntradiction.
If 0 (Z) = p3 = 0 (G) ⇒ Z = G => is abelian co ntradiction.
If 0 ( G ) = p2 ⇒ 0(G /E )
i.e ., G/Z is a group of prime order 'p' and so is cyclic ⇒ G is
obelian ⇒ contradiction ⇒ O(Z) ≠ p2 hence O(Z) = p]
Most Upvoted Answer
If p is a prime number and G is non-abelian group of order p3, then th...
Explanation:
To answer this question, we need to understand the properties of the center of a group and the structure of non-abelian groups of order p^3, where p is a prime number.
Center of a group:
The center of a group G, denoted as Z(G), is the set of elements that commute with every element of the group. In other words, for any element x in G, x is in Z(G) if and only if xg = gx for all g in G.
Non-abelian groups of order p^3:
Let G be a group of order p^3, where p is a prime number. If G is non-abelian, it means that G does not satisfy the commutative property, i.e., there exist elements x and y in G such that xy is not equal to yx.
Proof:
We will prove that the center of G has exactly p elements.
Step 1: The center of G is non-trivial.
Since G is non-abelian, there exists at least one element x in G such that there exists another element y in G where xy is not equal to yx. Let Z(G) be the center of G. Now, if x is not in Z(G), then there exists an element g in G such that xg is not equal to gx. But since G is of order p^3, the elements x, xg, gx, and g all belong to G. Therefore, Z(G) is non-empty.
Step 2: The center of G has at least p elements.
Let Z(G) have q elements, where q is less than p. Since Z(G) is non-trivial, there exists an element x in G that is not in Z(G). Now, consider the set {xg | g is in G}. This set contains p^3 elements since G is of order p^3. But since x is not in Z(G), xg is not equal to gx for at least one g in G. Therefore, the set {xg | g is in G} contains at least p^2 elements. Since q + p^2 is greater than p^3, we have a contradiction. Hence, Z(G) must have at least p elements.
Step 3: The center of G has at most p elements.
To prove this, we will assume that Z(G) has more than p elements. Let Z(G) have r elements, where r is greater than p. Now, consider the set {xg | x is in Z(G) and g is in G}. This set contains r*p elements. But since Z(G) has r elements, there must exist an element x in Z(G) such that the set {xg | g is in G} contains at least p elements. This means that xg is equal to gx for at least p elements of G. But since x is in Z(G), xg is equal to gx for all elements g in G. Therefore, G is abelian, which contradicts the given condition that G is non-abelian. Hence, Z(G) must have at most p elements.
Conclusion:
From Step 2 and Step 3, we can conclude that the center
To answer this question, we need to understand the properties of the center of a group and the structure of non-abelian groups of order p^3, where p is a prime number.
Center of a group:
The center of a group G, denoted as Z(G), is the set of elements that commute with every element of the group. In other words, for any element x in G, x is in Z(G) if and only if xg = gx for all g in G.
Non-abelian groups of order p^3:
Let G be a group of order p^3, where p is a prime number. If G is non-abelian, it means that G does not satisfy the commutative property, i.e., there exist elements x and y in G such that xy is not equal to yx.
Proof:
We will prove that the center of G has exactly p elements.
Step 1: The center of G is non-trivial.
Since G is non-abelian, there exists at least one element x in G such that there exists another element y in G where xy is not equal to yx. Let Z(G) be the center of G. Now, if x is not in Z(G), then there exists an element g in G such that xg is not equal to gx. But since G is of order p^3, the elements x, xg, gx, and g all belong to G. Therefore, Z(G) is non-empty.
Step 2: The center of G has at least p elements.
Let Z(G) have q elements, where q is less than p. Since Z(G) is non-trivial, there exists an element x in G that is not in Z(G). Now, consider the set {xg | g is in G}. This set contains p^3 elements since G is of order p^3. But since x is not in Z(G), xg is not equal to gx for at least one g in G. Therefore, the set {xg | g is in G} contains at least p^2 elements. Since q + p^2 is greater than p^3, we have a contradiction. Hence, Z(G) must have at least p elements.
Step 3: The center of G has at most p elements.
To prove this, we will assume that Z(G) has more than p elements. Let Z(G) have r elements, where r is greater than p. Now, consider the set {xg | x is in Z(G) and g is in G}. This set contains r*p elements. But since Z(G) has r elements, there must exist an element x in Z(G) such that the set {xg | g is in G} contains at least p elements. This means that xg is equal to gx for at least p elements of G. But since x is in Z(G), xg is equal to gx for all elements g in G. Therefore, G is abelian, which contradicts the given condition that G is non-abelian. Hence, Z(G) must have at most p elements.
Conclusion:
From Step 2 and Step 3, we can conclude that the center
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Community Answer
If p is a prime number and G is non-abelian group of order p3, then th...
Centre of G is 
Since 0(G) = p3, p is a prime number
But Z is a subgroup of G => 0 ( Z ) / 0 ( G ) i.e. 0 ( Z )
divides 0(G) ⇒ 0 (Z) / p3
⇒ 0 ( Z ) = p or p2 or p3.
If 0 (Z) = p3 = 0 (G) ⇒ Z = G => is abelian co ntradiction.
If 0 (Z) = p3 = 0 (G) ⇒ Z = G => is abelian co ntradiction.
If 0 ( G ) = p2 ⇒ 0(G /E )
i.e ., G/Z is a group of prime order 'p' and so is cyclic ⇒ G is
obelian ⇒ contradiction ⇒ O(Z) ≠ p2 hence O(Z) = p]
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If p is a prime number and G is non-abelian group of order p3, then the centre of G hasa)Exactly p - 1 elementsb)Exactly p elementsc)Exactly p3 elementd)None of theseCorrect answer is option 'B'. Can you explain this answer?
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