# Group center properties

1. Dec 2, 2011

### Menelaus

I am aware of the theorem |G/Z(G)|=p with p prime implies G/Z(G) is cyclic and thus G is abelian, but I do not understand why. Is there not a theorem that says G abelian $\Leftrightarrow$ Z(G)=G? So what if |G|=p$^{3}$ and |Z(G)|=p$^{2}$? This implies |G/Z(G)|=p implying G is abelian however G$\neq$Z(G). What is the ambiguity here?

2. Dec 2, 2011

### micromass

There is no ambiguity. The situation you mention can just never occur. You have just proven that it can never oocur that |G|=p3 and |Z(G)|=p2.

You can use this, for example to prove that all groups of order p2 are abelian. Indeed, Z(G) is a subgroup of G, so there are three possibilities:
• |Z(G)|=1 It can be proven that this can never happen (a group with prime order must always have nontrivial center)
• |Z(G)|=p Then your theorem says the group is abelian, thus |Z(G)|=p2, which is a contradiction.
• |Z(G)|=p2 is the only remaining possibility.

3. Dec 2, 2011

### Menelaus

Ah I see! Thanks for clearing that up.

4. Dec 3, 2011

### lavinia

Try looking at the exact sequence of groups,

0 -> Z(G) -> G -> G/Z(G) -> 1

- Any group of prime order is cyclic so G/Z(G) is cyclic.
- If the sequence is split then the group is isomorphic to the direct product Z(G) x G/Z(G)
so G is abelian and Z(G) = G
- If the sequence is not split what happens?

5. Dec 3, 2011

### Menelaus

I do not follow, I don't believe I've ever looked at this sequence or talked about what a split sequence is. Micromass cleared up the problem I was having though.

6. Dec 3, 2011

### Deveno

lavina, do you mean not left-split (because right-splits need not be direct products)?

7. Dec 5, 2011

### lavinia

To me split means that there is an inverse of the projection onto G/Z(G)

Since Z(G) is in the center of the group split means direct product - I think.

What you get is that there is a subgroup isomorphic to G/Z(G) that commutes with every other element of the group. this should be a direct product.

If the sequence is not split then a lift of a generator of the quotient has some power that lies in Z(G) but again it commutes with everything so G is abelian.

Last edited: Dec 5, 2011
8. Dec 5, 2011

### lavinia

I am sorry. I thought he only cleared up a special case. Anyway. This is a different way of looking at the problem that I thought you might find interesting.

9. Dec 6, 2011

### Deveno

ah, so you DID mean right-split. let me assure you that right-split ≠ direct product.

consider the following short exact sequence:

0→Cn→Dn→Dn/Cn→0.

(where we map a generator of Cn, to a generating rotation of Dn).

define:

f:Dn/Cn→Dn by:

f(Cn) = 1
f(Cns) = s

then if p is the canonical projection: p(f(Cnx)) = Cnx.

but Dn is certainly not a direct product of abelian groups.

10. Dec 6, 2011

### lavinia

If The kernel is in the center of the group then split means direct product. Otherwise not necessarily.I think the proof is straight forward.

Last edited: Dec 6, 2011
11. Dec 7, 2011

### Deveno

i see how this is true, because if H acts on N by conjugation, and N is in Z(G), the action has to be trivial.

12. Dec 8, 2011

### lavinia

yes. that's it.