Math Help for Finite Cyclic Group & Subgroups

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In summary: It's not {e}, because if it were then g^r={e} and g would be an element of C_r. So C_r must not be cyclic, and g^r must be an element of C_r but not of G. So g is not an element of G.
  • #1
cauchys_pet
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hey! great to find such an informative website...
i'm an undergrad math student and have lots of problems with group theory.i hope you'll all help me enjoy group theory...
my teacher put forward these question last week and I've been breaking my head over them without much success :
1. let G be a finite cyclic group of order p^n, p being prime and n >=0. if H and K are subgroups of G then show that either H contains K or K contains H.
i started out supposing the contrary but i wonder if I'm on the right track. i don't think it helps. :confused:

2.if G is a group of order 30 show that G has atmost 7 distinct subgroups of order 5.
can i say this : let H be a subgroup of order 5 then the number of distinct left cosets of H in G is 6. so are we done?!

3.let G be a group such that intesection of all subgroups of G different from {e}. then prove that every element of G has finite order.

4. give an example to show that a subgroup of index 3 may not be a normal subgroup of G. :frown:
thanks again for the help.
 
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  • #2
1. G is cyclic what does that mean? so what can you say about the elements in H and K in terms of this?


2. Is nothing to do with cosets. Suppose H and K are subgroups of order 5, then HnK is a subgroup whose order divides 5, so it follows HnK=H=K or HnK={e} So the subgroups are either equal or contain only one element in common. So if there 7 (or fewer) distinct subgroups of order 5 these contain 7*4+1=29 disticnt elements: they all contain e, and 4 other elements each that appear in exactly one subgroup. If there were more than 7 then what would happen?

3 makes no sense.

4. Hmm, can you think of any small subgroups that have a subgroup of index 3 that aren't abelian? Try the smallest such (it has order 6...)
 
  • #3
have been thinking about prob 3...i guess you take A6 (the set of even permutations of 6 elements), possibly find a subgroup H of order 4 and then look at the 3 distinct cosets of H.
 
  • #4
sorry, i goofed up problem number 2
it says, let G be a group such that intersection of all subgroups of G different from {e} is not {e}. prove that every element is of finite order.
 
  • #5
suppose there is an element of infinite order, g say.

For all r in N let C_r be the cyclic group generated by g^r,

What is the intersection of all these?
 

Related to Math Help for Finite Cyclic Group & Subgroups

1. What is a finite cyclic group?

A finite cyclic group is a group in which all elements can be generated by a single element, called a generator, by repeatedly applying the group operation. This means that the group is closed under the operation, and every element in the group can be expressed as a power of the generator.

2. How do you find the order of a finite cyclic group?

The order of a finite cyclic group is equal to the number of elements in the group. To find the order, you can count the number of elements that can be generated by the generator, or you can use the formula n/gcd(n,k), where n is the number of elements in the group and k is the order of the generator.

3. What is a subgroup of a finite cyclic group?

A subgroup of a finite cyclic group is a subset of the group that is also a group under the same operation. Subgroups can be generated by a single element, just like the original group, and can also be cyclic or non-cyclic.

4. How do you determine the subgroups of a finite cyclic group?

To determine the subgroups of a finite cyclic group, you can use the fact that every subgroup must contain the identity element and be closed under the group operation. This means that subgroups can be generated by any element that divides the order of the group.

5. Can a finite cyclic group have more than one subgroup of a given order?

Yes, a finite cyclic group can have multiple subgroups of the same order. For example, the group Z6 has two subgroups of order 2 (namely, {0,3} and {0,2,4}). However, there can be at most one cyclic subgroup of a given order, as every cyclic group can only have one generator of a given order.

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