Re: Groups of Finite Order

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**vidomagru** **1) Let G be a finite group of order 12, is it possible that the center of this group has order 4?**

I am not even sure how to approach this. I did have to look up the definition of the center of a group and I know that the center of a group G is defined as: $\displaystyle Z(G) =$ {$\displaystyle z \in G | \forall g \in G, zg=gz$}.

First a disclaimer: I have not taught any graduate level involving group theory is over twenty years, therefore I have not done any active work in this for that period. That said this may help you.

Any Abelian group is its own center. Thus you are looking for a non-Abelian group of order twelve that has a center of order four.

There has been quite a lot of work done on finite groups. If I were you that is where I would start.

Now there may very well be a completely clear solution that I don't see.

Re: Groups of Finite Order

There seem to be very few finite groups of order 12. Do I just have to go through each one or is possible to show this arbitrarily?

Re: Groups of Finite Order

Hi,

1. I leave it to you to prove the following fact:

If G is any non-abelian group (finite or infinite), the factor group G/Z(G) is not cyclic. If you have trouble, post your problems.

Now for a group G of order 12 with |Z(G)| = 4. Is G/Z(G) cyclic?

2. Unless you mean each p_{i} is a __power__ of a prime, you have misquoted the fundamental theorem. What you say is true about a finite abelian group G having a subgroup of order n for any divisor n of the order of G, but if you haven't proved this, I think it should be part of your solution. Alternatively, you can deduce the truth of the statement directly from the fundamental theorem, knowing only that subgroups of cyclic groups are cyclic.

Re: Groups of Finite Order

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Originally Posted by

**johng** Hi,

1. I leave it to you to prove the following fact:

If G is any non-abelian group (finite or infinite), the factor group G/Z(G) is not cyclic. If you have trouble, post your problems.

Now for a group G of order 12 with |Z(G)| = 4. Is G/Z(G) cyclic?

Ok, I think I can prove this: suppose $\displaystyle G/Z(G)$ is cyclic with generator $\displaystyle gZ, g \in G$. Let $\displaystyle a,b \in G$, then $\displaystyle a=g^nz, b=g^mz'$ for $\displaystyle z,z' \in Z(G)$ and $\displaystyle n,m \in \mathbb{Z}$. Since the center commutes with every elements of $\displaystyle G$, then $\displaystyle ab=g^nz \cdot g^mz' = g^{n+m}zz' = g^mz' \cdot g^nz = ba$, contradicting that G is nonabelian. Hence $\displaystyle G/Z(G)$ is not cyclic.

I would gather from my proof that if $\displaystyle |Z(G)| = 4$, then $\displaystyle G/Z(G)$ is not cyclic. But I am not sure I understand the implications of this.

Re: Groups of Finite Order

For G of order 12 and Z(G) of order 4, the order of G/Z(G) is 3. What do you know about any group of order 3 (or any prime)? Isn't such a group cyclic?

Re: Groups of Finite Order

Quote:

Originally Posted by

**johng** For G of order 12 and Z(G) of order 4, the order of G/Z(G) is 3. What do you know about any group of order 3 (or any prime)? Isn't such a group cyclic?

Oh I think I understand. Let G be a finite group of order 12. Assume that |Z(G)| = 4. Since any group of prime order is cyclic and since |G| = 12, |Z(G)| = 4, then we can say |G/Z(G)| = 3. Hence G/Z(G) is cyclic. However if G/Z(G) is cyclic then G is abelian. And since G is abelian if and only if Z(G) = G then |Z(G)| must equal 12, and therefore our assumption that |Z(G)|=4 is false, and G cannot have a center of order 4.

Is that right?

Re: Groups of Finite Order

Yes. By the way, good job on proving G/Z(G) is not cyclic unless G is abelian.