For a group G, |G|>6, it must have at least 4 conjugacy classes?

In summary, the question is whether there is a conjugacy class of size greater than or equal to 2 in a group of size |G|.
  • #1
lus1450
40
1

Homework Statement


Let [itex]G[/itex] be a group such that [itex]|G|>6[/itex]. Then there are at least 4 conjugacy classes in G

Homework Equations


The Attempt at a Solution


Well, I tried by contradiction by using the group of order 7, which must be isomorphic to the cyclic group of order 7, which has 7 conjugacy classes, but I feel that's VERY lacking.
 
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  • #2
You made a good start. You can now note that every group of prime order satisfies your theorem.

I don't know any overwhelming theorems about this, so if I were tackling this, I'd look next at p-groups, those whose order is a power of p, where p is prime. We have some useful theorems: a group of order ##p^2## is abelian. And the centralizer Z of any p-group is non-trivial. G/Z is a group. You can do something with this. Start by looking at groups of order 8 = ##2^3##.

Next I'd tackle the more general groups which are not p groups. You can use the Sylow theorems to decompose these into smaller subgroups. For example, a group of order 15 must have subgroups of order 3 and 5. So why must a (the) group of order 15 have more than 4 conjugacy classes?
 
  • #3
Why group of order 7, must be isomorphic to the cyclic group of order 7? For example Klein group is not isomorphic to the cyclic group of order 4.
 
  • #4
LagrangeEuler said:
Why group of order 7, must be isomorphic to the cyclic group of order 7? For example Klein group is not isomorphic to the cyclic group of order 4.

Groups of prime order must be cyclic: the order of an element is factor of the order of the group.
 
  • #5
To address the actual problem:

Zaculus said:

Homework Statement


Let [itex]G[/itex] be a group such that [itex]|G|>6[/itex]. Then there are at least 4 conjugacy classes in G

By the orbit-stablizer theorem applied to the conjugacy action of [itex]G[/itex] on itself, the number of elements in a conjugacy class is a factor of [itex]|G|[/itex]. There is always a conjugacy class of size 1 containing the identity.

Thus the question reduces to showing that if [itex]|G| > 6[/itex] and a set of [itex]|G| - 1[/itex] elements is partitioned into one or two disjoint non-empty subsets then the size of at least one of those subsets is not a factor of [itex]|G|[/itex].

That seems a much more manageable proposition than considering special cases of prime factorizations of [itex]|G|[/itex].
 

1. What is a group G?

A group G is a mathematical structure consisting of a set of elements and an operation that combines any two elements to produce a third element in the set. The operation must be associative, have an identity element, and each element must have an inverse.

2. What does |G|>6 mean?

The symbol |G| represents the cardinality, or number of elements, in the group G. So, |G|>6 means that the number of elements in the group is greater than 6.

3. What does it mean for a group to have conjugacy classes?

In a group, elements that can be transformed into each other by an inner automorphism are said to be in the same conjugacy class. This means that they have the same essential algebraic properties, even though they may look different.

4. Why must a group with |G|>6 have at least 4 conjugacy classes?

This is a consequence of a theorem called the Class Equation, which states that the total number of elements in a group is equal to the sum of the size of the conjugacy classes plus the number of elements that are not in any conjugacy class. Since |G|>6, there must be at least 4 elements that are not in any conjugacy class, meaning that there must be at least 4 conjugacy classes.

5. Can you give an example of a group with |G|>6 and 4 conjugacy classes?

Yes, the group S4 (the symmetric group on 4 elements) has 24 elements and 4 conjugacy classes. The conjugacy classes are: the identity element, 6 elements of order 2, 8 elements of order 3, and 9 elements of order 4.

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