Number of Subgroups of a Group

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In summary, the number of subgroups of a group is a useful mathematical concept that helps us understand the structure and properties of a group. It can be calculated by finding all distinct combinations of elements and checking if they form a subgroup, and can be simplified using mathematical theorems and formulas. The number of subgroups can be infinite for infinite groups, but is always a finite number for finite groups. It is closely related to the group's order, with the number of subgroups being a divisor of the order. The number of subgroups can change depending on the group's properties and structure, such as adding new elements or changing the group's operation.
  • #1
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if |G| = n . And G has 2n-1 subgroups, prove G = <e> or it is isomorphic to Z2.

I am thinking I need to show that n is either 1 or 2. But I don't know how to use the number of subgroups. I can't find a theorem that mentions the number of subgroups or a group.

any Hints please?
 
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  • #2
I guess I was looking at this the wrong way. I thought this would be an application of Lagrange thm, but all I need is the fact that if G has 2n-1 subgroups, then G has at least 2n-1 elements, hence n ≥ 2n-1

Which is only true if n =1 or n= 2.
 
  • #3
This got me thinking, if Z = <1>, is my proof still correct in the case of G = <e>. If the group operation was addition. But I wasn't given any operation.
Should it be that if G = <e>, then like Z, G may be infinite so G has infinite subgrps, and it is still true that n = ∞ ≥ 2n-1.

What do you think?
 
  • #4
you are given that |G| = n. presumably, then, G is finite.
 
  • #5
Deveno said:
you are given that |G| = n. presumably, then, G is finite.

Hi Deveno. It's refreshing to see you around. :)

true, hence my early proof is correct, and G = <e> if |G|=1. right?
 
  • #6
Bachelier said:
the fact that if G has 2n-1 subgroups, then G has at least 2n-1 elements

Why do you think this is true? Z2 x Z2 has 5 subgroups, but only 4 elements. Thus, your earlier proof doesn't appear correct.As for how to prove it, I think a simple counting argument is enough. If |G| = n, let N denote the number of subgroups of G. If d is a positive integer, then there are at most [itex]\binom{n-1}{d-1}[/itex] subgroups of G of order d (since the identity must be in the subgroup, and there are d-1 elements to choose out of the remaining n-1). Now, if there is a subgroup of order d, then d divides n by Lagrange, so either d = n or 1 ≤ d ≤ n/2. Thus, the number of subgroups of G satisfies
[tex]N \le \binom{n-1}{n-1} + \sum_{d=1}^{\lfloor n/2 \rfloor} \binom{n-1}{d-1} \le 1 + \sum_{k=0}^{\lfloor n/2 \rfloor - 1} \binom{n-1}{k} \le 1 + 2^{n-2}.[/tex]
The last inequality follows from the fact that [itex]\sum_{k=0}^{\lfloor n/2 \rfloor - 1} \binom{n-1}{k}[/itex] consists of at most the first half of the terms in [itex]\sum_{k=0}^{n-1} \binom{n-1}{k} = 2^{n-1}[/itex], and the fact that [itex]\binom{n-1}{k} = \binom{n-1}{n-1-k}[/itex].

Now if N = 2n-1, then N ≤ 1 + 2n-2 can only hold if n = 1 or 2.
 
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  • #7
it seems of interest to show that if you have two subgroups that are not nested, then their union is probably not a subgroup. that should do it.
 
  • #8
mathwonk said:
it seems of interest to show that if you have two subgroups that are not nested, then their union is probably not a subgroup. that should do it.

It is clear the union (unlike the intersection) is not a subroup. In this case if H1 is the nesting set, it is the union, hence it is not a subgroup contradicting our assumption.

But isn't this a special case? Where do we use 2n-1
 
  • #9
see, i would be inclined to use induction, like so:

if G is a group of n elements, with n > 2, then there is some subset containing {e} of G that is not a group.

if |G| = 3, then {e,a} for any element a ≠ e, will work.

assume true for any group of order 2 < k < n.

now consider |G| = n. choose some non-identity element x in G. there are two cases:

G = <x>, or G ≠ <x>.

if G ≠ <x>, we can apply our induction hypothesis to <x>, and we're done.

if n is not prime, say n = kd, then choose <xd>, and we're done.

so it suffices to consider G cyclic of prime order. since n > 2, n = p is an odd prime.

hence {e,x} is not a group.

how many subsets are there, of the form {e} U X, where X ⊆ G-{e}?
 
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  • #10
Deveno said:
if G is a group of n elements, with n > 2, then there is some subset containing {e} of G that is not a group.

if |G| = 3, then {e,a} for any element a ≠ e, will work.

Hi Deveno.
Why is it that if G is a group of n elements, with n > 2, then there is some subset containing {e} of G that is not a group?
I see for the case of n=3 but how to generalize it. Actually in the case |G|= 3, G has only 2 subgroups (or 3 if you include itself). Actually I never thought about that: is G a subgroup of G?
 
  • #11
Bachelier said:
It is clear the union (unlike the intersection) is not a subroup. In this case if H1 is the nesting set, it is the union, hence it is not a subgroup contradicting our assumption.

But isn't this a special case? Where do we use 2n-1

Didn't notice that you were talking about the disjoint union. Still easy to prove, then what?
 
  • #12
yes, G is a subgroup of G (not a proper subgroup, however).

my previous post is a proof of that assertion (that if |G| > 2, G has a subset containing {e} which is not a subgroup), using induction on |G|. if you do not understand the proof, i can try to clarify.
 

1. What is the purpose of calculating the number of subgroups of a group?

The number of subgroups of a group is a useful mathematical concept that helps us understand the structure and properties of a group. It can also provide insights into other areas of mathematics and help solve complex problems.

2. How is the number of subgroups of a group calculated?

The number of subgroups of a group can be calculated by finding all the distinct combinations of elements in a group and checking if they form a subgroup. This process can be simplified using certain mathematical theorems and formulas.

3. Can the number of subgroups of a group be infinite?

Yes, the number of subgroups of a group can be infinite. This is especially true for infinite groups, where the number of subgroups is also infinite. However, for finite groups, the number of subgroups is always a finite number.

4. How does the number of subgroups of a group relate to its order?

The number of subgroups of a group is closely related to its order. In fact, the number of subgroups of a group is always a divisor of the order of the group. This relationship can help in determining the number of subgroups for a given group.

5. Can the number of subgroups of a group change?

Yes, the number of subgroups of a group can change depending on the group's properties and structure. For example, if a new element is added to the group, the number of subgroups may increase or decrease. Similarly, if the group's operation is changed, the number of subgroups may also change.

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