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Let the order of group G is 47, then find out the total no. of proper subgroup.
    Correct answer is '0'. Can you explain this answer?
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    Let the order of group G is 47, then find out the total no. of proper ...
    Since 47 is a prime no. i. e. group G has no proper subgroup.
    Note: It is a well known theorem that Every group of prime order is a simple group and simple group has no proper normal subgroup.
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    Let the order of group G is 47, then find out the total no. of proper ...
    Order of given group is 47 which is prime, so 47 =1into 47 ..so this group will contain only two subgroups of orders 1 and 47 each..which means those subgroups will be of the form
    1)subgroup containing only identity element and
    2)group itself..
    which are not proper subgroups.(subgroups other than 1) and 2) are referred to as proper subgroups)
    so answer will be zero.
    i.e. Total no. of proper subgroups in group G is Zero.
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    Let the order of group G is 47, then find out the total no. of proper ...
    Since 47 is a prime no. i. e. group G has no proper subgroup.
    Note: It is a well known theorem that Every group of prime order is a simple group and simple group has no proper normal subgroup.
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    Let the order of group G is 47, then find out the total no. of proper subgroup.Correct answer is '0'. Can you explain this answer?
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