# Non-abelian group or order 27

1. Jun 15, 2007

### Ultraworld

An example in my book on character theory gives an arbitrary non-abelian group G of order 27 and claims that |G'| = 3 (order of the commutator subgroup equals 3) and that there are 11 conjugation-classes. I really don't know why. I only manage to deduce that |Z(G)| = 3 (the order of the center of the group equals 3).

Last edited: Jun 15, 2007
2. Jun 18, 2007

### CompuChip

I don't know exactly (haven't really given it deep thought) but...
- The commutator subgroup is a subgroup, so, it has to have order 1, 3, 9 or 27 (divides |G| = 27)
- It cannot be 1. Then the commutator subgroup would be trivial (every commutator is unity) and the group would be abelian.
- You already deduced that |Z(G)| = 3, so there are three elements commuting with everything. Therefore, it cannot be 27 either.
- So you still have to exclude the case 9.

Probably there is some property (or Sylow theorem that can give you that. Hope that gives you a nudge in the right direction (and I'm not putting you on a wrong track here)

3. Jun 18, 2007

### StatusX

The commutator subgroup has the property that it is the smallest normal subgroup H such that the quotient G/H is abelian, in the sense that any other subgroup K with G/K abelian has H<K. Now just use the fact you've shown about the center (which, remember, is normal).

4. Jun 19, 2007

### Ultraworld

|Z(G)| = 3 therefore |G/Z(G)| = 9 therefore (groups of order p2 are abelian) G/Z(G) is abelian therefore Z(G) > G' therefore Z(G) = G' therefore |G'| = 3.

Last edited: Jun 19, 2007
5. Jun 28, 2007

### Ultraworld

I still do not know why there are 11 conj-classes.

Im afraid I just do not know enough small tiny theorems to prove it. I only know

1. The identity element is in a conj-classs of its own.
2. The order of a conj-class devides the group order.

So if I could prove there are no conj-class or order 9 then Im done. But I do not know how to prove that.

Last edited: Jun 28, 2007
6. Jun 28, 2007

### Kummer

The Class equation for G is:
$$|G|=|Z(G)| + \sum_{k=s+1}^n |Gx_k|$$
The orbits $$|Gx_k|>1$$ and are divisors of |G|=27.
So $$|Gx_k|$$ are either 3 or 9.

Now it cant be 11 cojugate classes.
Because even in the smallest possible case where $$|Gx_k|=3$$ we end up with,
27 = 3 + (3+3+..+3)
Now the number of 3's is 9.
So the most number of conjugate classes possible is 9.

7. Jun 28, 2007

### StatusX

That's almost right. The 3 elements in the center each form their own conjugacy classes, so in fact the maximum number of conjugacy classes is 11.

The class equation is the way to go, if you're familiar with it. Note that if an element g is not in the center, its centralizer C_G(g) is a proper subgroup which contains both g and the center, therefore Z<C_G(g)<G, where both inclusions are proper.

8. Jun 28, 2007

### mathwonk

there is only one elementary principle in group theory, namely the group acts on its elements, subgroups, cosets, subsets, etc.....

and the kernel of such an action is a normal subgroup.

more precisely there are two such actions, translation, and conjugation. playing with these gives everything else, like the "class" formulas above.

well there is also a more elementary principle, namely counting elements.

9. Jun 29, 2007

### Kummer

I made a mistake in adding, the class equation gives:
27 = (1 + 1 + 1) + (3+3+3+3+3+3+3+3)

So there are 11 conjugacy classes.

10. Jul 1, 2007

### Ultraworld

ok here we go

Take arbitrary g in G. We define the subgroup C(g) = {x in G : xg = gx}. We got

Z(G) < C(g) < G.

So |C(g)| = 3, 9, 27. Now assume g not in Z(G). We make case distinction

• order 3 is impossible cause g in C(g) but g was not in Z(G).
• order 9, we leave this one empty.
• order 27. That would mean C(g) = G but g was not in Z(G) so impossible

We conclude |C(g)| = 9. So there exists x and y in G such that

G = C(g) U xC(g) U yC(g).

Take an arbitrary element c in C(g). Than cgc-1 = g. (by definition)
Take an arbitrary element xc in C(g). Than (xc)g(xc)-1 = xcgc-1x-1 = xgx-1.
Take an arbitrary element yc in C(g). Than (yc)g(yc)-1 = ycgc-1y-1 = ygy-1.

So the conjugacy class of g consist of just 3 elements.

We now make our final conclusion. Because the order of a conjugacy class devides |G| there are 3 conjugacy classes of order 1 (|Z(G)| = 3) and 8 of order 3:

1 + 1 + 1 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = 27.

So a total of 11 conjugacy classes.

Last edited: Jul 1, 2007
11. Jul 1, 2007

### Ultraworld

Thanks for the help guys. Now try wetter I can do this for a non-abelian group of order 48. Because that is the real exercise. This was just a example in the book (but they did not explained why they got this all)

Last edited: Jul 1, 2007
12. Jul 1, 2007

### Ultraworld

I suspect the following proposition holds

proposition: Given arbitrary g in G. Then the order of the conjugacy class which contains g equals [G : C(g)].

edit: http://en.wikipedia.org/wiki/Conjugacy_class says Im right. :-)

Now Im happy cause I do have some structure for that non-abelian group of order 48.

Last edited: Jul 1, 2007
13. Jul 1, 2007

### Ultraworld

proof of proposition:
Given arbitrary g in G. Given x and y in G such that xC(g) != yC(g). So C(g) != x-1yC(g) so x-1y not in C(g). Now assume

xgx-1 = ygy-1.

We easily derive from this

g = (x-1y)g(x-1y)-1.

So x-1y in C(g). Contradiction so

xgx-1 != ygy-1.

so the order of the conjugacy class of g equals [G : C(g)].

Last edited: Jul 1, 2007
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