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If K is proper subgroup of H and H is a proper subgroup of G , where O(k) = 35 and 0(G) = 700, then how many subgroups H are possible ?
Correct answer is '4'. Can you explain this answer?
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If K is proper subgroup of H and H is a proper subgroup of G , where O...
Let 0(H) = m then by lagrange's theorem 0(H) | 0(G) and 0(K) | 0(H).
i.e. 35|m and m | 700
Thus m is multiple of 35 and divisor of 700.
Now 700 = 35 x 20
So, the possible values of m are 35, 2 x 3 5 , 4 x 3 5 , 5 x 3 5 , 10 x 3 5 , 20x35, i.e. 3 5 ,7 0 ,140 ,175 ,350 ,700 .
But K is proper subgroup of H, m ≠ 35 and H is proper subgroup of G, m ≠ 700.
Hence the possible orders of H are 70 ,140,175 and 350 total possible H are 4
Most Upvoted Answer
If K is proper subgroup of H and H is a proper subgroup of G , where O...
To find the number of possible subgroups H, we need to consider the order of H. The order of a subgroup is defined as the number of elements in the subgroup.
Given that K is a proper subgroup of H, we know that the order of K is a divisor of the order of H. Similarly, since H is a proper subgroup of G, the order of H is a divisor of the order of G.
O(K) = 35 and O(G) = 700, so let's consider the possible divisors of 35 and 700 to find the order of H.
Prime factorization of 35: 5 * 7
Prime factorization of 700: 2^2 * 5^2 * 7
To find the order of H, we need to select some or all of the prime factors from the prime factorization of G, excluding any that are already present in K.
Considering the prime factors without repetition, we have the following possibilities for the order of H:
1. Selecting only the prime factor 2: The order of H would be 2, but this is not possible since K is a proper subgroup of H, and the order of K is 35, which is not divisible by 2.
2. Selecting only the prime factor 5: The order of H would be 5, but this is also not possible since K is a proper subgroup of H, and the order of K is 35, which is not divisible by 5.
3. Selecting only the prime factor 7: The order of H would be 7, and this is possible since K is a proper subgroup of H, and the order of K is 35, which is divisible by 7.
4. Selecting the prime factors 2 and 5: The order of H would be 10, but this is not possible since K is a proper subgroup of H, and the order of K is 35, which is not divisible by 10.
Therefore, there are 2 possible subgroups H: one with order 7 and one with order 35.
However, we also need to consider the case where we select all the prime factors from the prime factorization of G, excluding any that are already present in K. In this case, the order of H would be 2^2 * 5^2 * 7, which is equal to 140.
So, in total, there are 3 possible subgroups H: one with order 7, one with order 35, and one with order 140.
However, we need to consider the fact that H is a proper subgroup of G, which means its order must be less than the order of G. Since the order of G is 700, the subgroup H with order 140 is not possible.
Therefore, the correct answer is 4 possible subgroups H: one with order 7, one with order 35, and two with order less than 35.
Given that K is a proper subgroup of H, we know that the order of K is a divisor of the order of H. Similarly, since H is a proper subgroup of G, the order of H is a divisor of the order of G.
O(K) = 35 and O(G) = 700, so let's consider the possible divisors of 35 and 700 to find the order of H.
Prime factorization of 35: 5 * 7
Prime factorization of 700: 2^2 * 5^2 * 7
To find the order of H, we need to select some or all of the prime factors from the prime factorization of G, excluding any that are already present in K.
Considering the prime factors without repetition, we have the following possibilities for the order of H:
1. Selecting only the prime factor 2: The order of H would be 2, but this is not possible since K is a proper subgroup of H, and the order of K is 35, which is not divisible by 2.
2. Selecting only the prime factor 5: The order of H would be 5, but this is also not possible since K is a proper subgroup of H, and the order of K is 35, which is not divisible by 5.
3. Selecting only the prime factor 7: The order of H would be 7, and this is possible since K is a proper subgroup of H, and the order of K is 35, which is divisible by 7.
4. Selecting the prime factors 2 and 5: The order of H would be 10, but this is not possible since K is a proper subgroup of H, and the order of K is 35, which is not divisible by 10.
Therefore, there are 2 possible subgroups H: one with order 7 and one with order 35.
However, we also need to consider the case where we select all the prime factors from the prime factorization of G, excluding any that are already present in K. In this case, the order of H would be 2^2 * 5^2 * 7, which is equal to 140.
So, in total, there are 3 possible subgroups H: one with order 7, one with order 35, and one with order 140.
However, we need to consider the fact that H is a proper subgroup of G, which means its order must be less than the order of G. Since the order of G is 700, the subgroup H with order 140 is not possible.
Therefore, the correct answer is 4 possible subgroups H: one with order 7, one with order 35, and two with order less than 35.
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If K is proper subgroup of H and H is a proper subgroup of G , where O(k) = 35 and 0(G) = 700, then how many subgroups H are possible ?Correct answer is '4'. Can you explain this answer?
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If K is proper subgroup of H and H is a proper subgroup of G , where O(k) = 35 and 0(G) = 700, then how many subgroups H are possible ?Correct answer is '4'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about If K is proper subgroup of H and H is a proper subgroup of G , where O(k) = 35 and 0(G) = 700, then how many subgroups H are possible ?Correct answer is '4'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If K is proper subgroup of H and H is a proper subgroup of G , where O(k) = 35 and 0(G) = 700, then how many subgroups H are possible ?Correct answer is '4'. Can you explain this answer?.
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