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If a cyclic group has only 3 subgroups {e}, a subgroup of order 7 and the group itself. Then, the order of group is
- a)7
- b)14
- c)49
- d)cannot be determined
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
If a cyclic group has only 3 subgroups {e}, a subgroup of order 7 and ...
To determine the order of the cyclic group, we need to analyze its subgroups.
Cyclic Group and Subgroups
A cyclic group is a group generated by a single element. Let's denote the generator of the cyclic group as "g". Since the cyclic group has only three subgroups, let's analyze each subgroup to understand its properties.
1. {e} Subgroup
The subgroup {e} consists of only the identity element. Every group contains this subgroup, so it doesn't provide any additional information about the order of the cyclic group.
2. Subgroup of Order 7
The subgroup of order 7 indicates that there is an element within the cyclic group that has an order of 7. Let's denote this element as "g^7". Since the subgroup has 7 elements, it means that "g^7" generates all these elements. Therefore, the order of the cyclic group must be a multiple of 7.
3. The Group Itself
The third subgroup is the cyclic group itself. Since a group is always a subgroup of itself, this doesn't provide any new information about the order of the cyclic group.
Determining the Order
From the given information, we know that the cyclic group has a subgroup of order 7. This indicates that the order of the cyclic group must be a multiple of 7. However, we don't have any additional information to determine the exact order.
For example, the cyclic group could have an order of 7, where the subgroup of order 7 is the entire group itself. On the other hand, the cyclic group could have an order of 49, where the subgroup of order 7 is a proper subgroup within the group.
Therefore, the order of the cyclic group cannot be determined solely based on the given information. The correct answer is option C, "cannot be determined."
Cyclic Group and Subgroups
A cyclic group is a group generated by a single element. Let's denote the generator of the cyclic group as "g". Since the cyclic group has only three subgroups, let's analyze each subgroup to understand its properties.
1. {e} Subgroup
The subgroup {e} consists of only the identity element. Every group contains this subgroup, so it doesn't provide any additional information about the order of the cyclic group.
2. Subgroup of Order 7
The subgroup of order 7 indicates that there is an element within the cyclic group that has an order of 7. Let's denote this element as "g^7". Since the subgroup has 7 elements, it means that "g^7" generates all these elements. Therefore, the order of the cyclic group must be a multiple of 7.
3. The Group Itself
The third subgroup is the cyclic group itself. Since a group is always a subgroup of itself, this doesn't provide any new information about the order of the cyclic group.
Determining the Order
From the given information, we know that the cyclic group has a subgroup of order 7. This indicates that the order of the cyclic group must be a multiple of 7. However, we don't have any additional information to determine the exact order.
For example, the cyclic group could have an order of 7, where the subgroup of order 7 is the entire group itself. On the other hand, the cyclic group could have an order of 49, where the subgroup of order 7 is a proper subgroup within the group.
Therefore, the order of the cyclic group cannot be determined solely based on the given information. The correct answer is option C, "cannot be determined."
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If a cyclic group has only 3 subgroups {e}, a subgroup of order 7 and ...
Assume that a cyclic group G has exactly three subgroups; G itself{e}, and a subgroup of order 7.
The objective is to find |G|
Give the conclusion if 7 is replaced by prime number p.
By the hypothesis, G is a cyclic group, so there is an element a belongs to G such that =G
Since G has a subgroup of order 7, so 7 divides the order G.
That is,
=|G|=7n
For n=1
The cyclic group G and one of its subgroups have same order n.
By the Fundamental Theorem of Cyclic Groups,
“Every subgroup of a cyclic group is cyclic, Moreover, if<.a>=n , then the order of any subgroup of is a divisor of n; and, for each positive divisor k of n , the group has exactly one subgroup of order k , namely, ”.
Since G has exactly three subgroups,G, {e} and a subgroup of order 7, obtain has exactly three divisors 1, 7 and n.
For the case of n=1 the cyclic group G and subgroup of order 7 are the same.
This contradicts the fact that, G has 3 distinct subgroups.
For n=2,3,4,5,6
By the Fundamental Theorem of Cyclic Groups G= , has subgroup of order n.
So the cyclic group G has at least 4 subgroups: {e} G itself, the subgroup of order 7, and the sub group of order n.
This contradicts the fact that, G has 3 distinct subgroups.
For
n=7
|G|=7.7=49
Here 7 is the only positive divisor of 49 except 1 and 49.
That is, 7 is the only possible order of a subgroup other than{e} or G.
By the Fundamental Theorem of Cyclic Groups, there is exactly one subgroup pf order 7.
So for n<=7, g="" has="" a="" subgroup="" of="" order="">
That is, G has at least 4 distinct subgroups.
Hence, it is concluded that,
|G|=7^2=49
If 7 is replaced with p, where p is a prime, obtain that n has exactly three divisors 1, p and n.
Because G has exactly three subgroups,G,{e} and a subgroup of order p.
Hence, the order of p is, n=p^2
The objective is to find |G|
Give the conclusion if 7 is replaced by prime number p.
By the hypothesis, G is a cyclic group, so there is an element a belongs to G such that =G
Since G has a subgroup of order 7, so 7 divides the order G.
That is,
=|G|=7n
For n=1
The cyclic group G and one of its subgroups have same order n.
By the Fundamental Theorem of Cyclic Groups,
“Every subgroup of a cyclic group is cyclic, Moreover, if<.a>=n , then the order of any subgroup of is a divisor of n; and, for each positive divisor k of n , the group has exactly one subgroup of order k , namely,
Since G has exactly three subgroups,G, {e} and a subgroup of order 7, obtain has exactly three divisors 1, 7 and n.
For the case of n=1 the cyclic group G and subgroup of order 7 are the same.
This contradicts the fact that, G has 3 distinct subgroups.
For n=2,3,4,5,6
By the Fundamental Theorem of Cyclic Groups G=
So the cyclic group G has at least 4 subgroups: {e} G itself, the subgroup of order 7, and the sub group of order n.
This contradicts the fact that, G has 3 distinct subgroups.
For
n=7
|G|=7.7=49
Here 7 is the only positive divisor of 49 except 1 and 49.
That is, 7 is the only possible order of a subgroup other than{e} or G.
By the Fundamental Theorem of Cyclic Groups, there is exactly one subgroup pf order 7.
So for n<=7, g="" has="" a="" subgroup="" of="" order="">
That is, G has at least 4 distinct subgroups.
Hence, it is concluded that,
|G|=7^2=49
If 7 is replaced with p, where p is a prime, obtain that n has exactly three divisors 1, p and n.
Because G has exactly three subgroups,G,{e} and a subgroup of order p.
Hence, the order of p is, n=p^2
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If a cyclic group has only 3 subgroups {e}, a subgroup of order 7 and the group itself. Then, the order of group isa)7b)14c)49d)cannot be determinedCorrect answer is option 'C'. Can you explain this answer? for Mathematics 2025 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about If a cyclic group has only 3 subgroups {e}, a subgroup of order 7 and the group itself. Then, the order of group isa)7b)14c)49d)cannot be determinedCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mathematics 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a cyclic group has only 3 subgroups {e}, a subgroup of order 7 and the group itself. Then, the order of group isa)7b)14c)49d)cannot be determinedCorrect answer is option 'C'. Can you explain this answer?.
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