Thread: Groups of Finite Order - IGNORE THIS - MOVED TO ALG.

1. Groups of Finite Order - IGNORE THIS - MOVED TO ALG.

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

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 $\displaystyle \mathbb{Z}_p_1 \oplus \mathbb{Z}_p_2 \oplus ... \oplus \mathbb{Z}_p_n$ where $\displaystyle p_n$ are not necessarily distinct and where $\displaystyle |G| = p_1 \cdot p_2 \cdot \cdot \cdot p_n$. And a corollary follows stating that if $\displaystyle m$ divides the order of a finite abelian group $\displaystyle G$, then $\displaystyle G$ has a subgroup of order $\displaystyle m$.

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 $\displaystyle H$ has order $\displaystyle 42 = 2 \cdot 3 \cdot 7$, we see that $\displaystyle H \simeq \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_7 \simeq \mathbb{Z}_42$. Since $\displaystyle \mathbb{Z}_p$ is cyclic for any prime $\displaystyle p$, $\displaystyle H$ is cyclic. $\displaystyle \blacksquare$

Does that work?

2. Re: Groups of Finite Order - IGNORE THIS - MOVED TO ALG.

Posted this in the wrong section... oops... I'll move it to advanced algebra