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

[lus1450]

Homework Statement

Let $G$ be a group such that $|G|>6$. 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.

 

[brmath]

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?

 

[LagrangeEuler]

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.

 

[pasmith]

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.

 

[pasmith]

To address the actual problem:

Homework Statement

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

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

Thus the question reduces to showing that if $|G| > 6$ and a set of $|G| - 1$ 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 $|G|$.

That seems a much more manageable proposition than considering special cases of prime factorizations of $|G|$.