# Groups of Finite Order

• Oct 28th 2013, 01:16 PM
vidomagru
Groups of Finite Order
I have two questions, one of which I am pretty sure I have answered, but I would like your opinion on it.

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:
$Z(G) =$ { $z \in G | \forall g \in G, zg=gz$}.

2) Suppose that the order of some finite Abelian group G is divisible by 42. Prove that G has a cyclic subgroup of order 42.
Let G be a finite abelian group of order 42. Remember that the Fundamental Theorem of Finite Abelian Groups states that any finite abelian group can be written as $\mathbb{Z}_p_1 \oplus \mathbb{Z}_p_2 \oplus ... \oplus \mathbb{Z}_p_n$ where $p_n$ are not necessarily distinct and where $|G| = p_1 \cdot p_2 \cdot \cdot \cdot p_n$. And a corollary follows stating that if $m$ divides the order of a finite abelian group $G$, then $G$ has a subgroup of order $m$.

Now since 42 divides the order of G, and G is abelian then we know that G contains a subgroup of order 42. Let H be such a subgroup. Since G is abelian, H is abelian. Since $H$ has order $42 = 2 \cdot 3 \cdot 7$, we see that $H \simeq \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_7 \simeq \mathbb{Z}_42$. Since $\mathbb{Z}_p$ is cyclic for any prime $p$, $H$ is cyclic. $\blacksquare$

Does that work?
• Oct 28th 2013, 07:19 PM
Plato
Re: Groups of Finite Order
Quote:

Originally Posted by 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: $Z(G) =$ { $z \in G | \forall g \in G, zg=gz$}.

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.
• Oct 29th 2013, 12:03 PM
vidomagru
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?
• Oct 29th 2013, 08:24 PM
johng
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 pi 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.
• Nov 1st 2013, 10:50 AM
vidomagru
Re: Groups of Finite Order
Quote:

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

I would gather from my proof that if $|Z(G)| = 4$, then $G/Z(G)$ is not cyclic. But I am not sure I understand the implications of this.
• Nov 1st 2013, 10:59 AM
johng
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?
• Nov 1st 2013, 12:09 PM
vidomagru
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?
• Nov 1st 2013, 07:06 PM
johng
Re: Groups of Finite Order
Yes. By the way, good job on proving G/Z(G) is not cyclic unless G is abelian.