Is N Always in the Center of G or Does It Intersect H or K Nontrivially?

[Bashyboy]

Homework Statement

Let $H, K, N$ be nontrivial normal subgroups of a group $G$ and suppose that $G = H \times K$. Prove that $N$ is in the center of $G$ or $N$ intersects one of $H,K$ nontrivially

Homework Equations

The Attempt at a Solution

I presume that $G = H \times K$ means that $G$ is the internal direct product of the subgroups of $H$ and $K$. So $G$ being the internal direct product of $G$ means that $G= \langle H \cup K \rangle$ with $H \cap K = \{e\}$; and since at least one of the two subgroups is normal, $\langle H \cup K \rangle = HK$ (Is this also called the internal weak direct product?) Now, suppose that $N$ intersects $H$ and $K$ trivially. Then $nh = hn$ for all $n \in N$ and $h \in H$ ; and a similar commutation relation holds for $K$. Then given $g \in HK$, which means $g = hk$ for some $h \in H$ and $k \in K$, we have that $nhk = hkn$, thereby showing $n \in Z(G)$.

Does this sound right?

 

[fresh_42]

Homework Statement

Let $H, K, N$ be nontrivial normal subgroups of a group $G$ and suppose that $G = H \times K$. Prove that $N$ is in the center of $G$ or $N$ intersects one of $H,K$ nontrivially

Homework Equations

The Attempt at a Solution

I presume that $G = H \times K$ means that $G$ is the internal direct product of the subgroups of $H$ and $K$. So $G$ being the internal direct product of $G$ means that $G= \langle H \cup K \rangle$ with $H \cap K = \{e\}$; and since at least one of the two subgroups is normal, ...

Aren't all already given normal?

... $\langle H \cup K \rangle = HK$ (Is this also called the internal weak direct product?) Now, suppose that $N$ intersects $H$ and $K$ trivially.

Which means, that we have to show $N \subseteq Z(G)$ or equivalently $nh=hn$ and $nk=kn$

Then

is what has to be shown. Simply writing it doesn't show anything. Why does it have to be the case? All we know for sure is $nh=h'n$.

$nh = hn$ for all $n \in N$ and $h \in H$ ; and a similar commutation relation holds for $K$. Then given $g \in HK$, which means $g = hk$ for some $h \in H$ and $k \in K$, we have that $nhk = hkn$, thereby showing $n \in Z(G)$.

Does this sound right?

 

[Bashyboy]

Sorry. I forgot to mention that I am using the following: if $K$ and $N$ are normal in $G$ and $K \cap N = \{e\}$, then $nk=kn$ for all $k \in K$ and $n \in N$. This I have already proven.

 

[Bashyboy]

I'm pretty certain the above is right. The second part of the problem is to find a nonabelian $G$ and normal subgroups $N,H,K$ such that $G = H \times K$ and $N$ contains the center. At first I thought about $D_8$, but the orders of the normal subgroups don't work out. I could a hint as to where I should be looking for an example. The only nonabelian groups I know of at this point are $S_n$, $D_{2n}$, and the quaternions.

 

[fresh_42]

Why don't simply choose $N=K=Z(G)$ and any group $H$ with $Z(H)=1\,$?

 

[Bashyboy]

Why don't simply choose $N=K=Z(G)$ and any group $H$ with $Z(H)=1\,$?

This is something I had in mind, but I didn't know what specific groups to look at. By the way, $H$ can't be any group; It has to be a normal subgroup of $G$, right?

 

[Bashyboy]

Ah I think I see what you are aiming at. Also, in order to dispel a possible misconception, when I write $G = H \times K$, this denotes the internal direct product (not a cartesian product); I seem to have adopted this horrible notation from Hungerford (the book I am working through). In fact, I am going to set up a notational convention right now: $H \times K$ will denote the standard direct (cartesian) product of $H$ and $K$, while $H \times_w K$ will denote the internal weak direct product (i.e., $G=HK$ and $H,K$ are normal subgroups with trivial intersection). I hope I am using all these terms correctly.

I believe you are aiming at the following: If $G \simeq G_1 \times G_2$, then there exist normal subgroups $H,K$ of $G$ such that $G = H \times_w K$ and $G_1 \simeq H$ and $G_2 \simeq K$. Here is my proof:

Let $f : G_1 \times G_2 \to G$ be an isomorphism. Note that $G_1 \times \{e_2\}$ and $\{e_1\} \times G_2$ are normal subgroups of $G_1 \times G_2$, and therefore $H := f(G_1 \times \{e_2\})$ and $K := f(\{e_1\} \times G_2)$ are normal subgroups of $G$. Suppose that $g \in H \cap K$. Then $f(g_1,e_2) = g = f(e_1,g_2)$ and injectivity of $f$ implies $g_1 = e_1$ and $g_2 = e_2$, whence it follows $g=e$. Now suppose that $g \in G$. Then for some $(g_1,g_2) \in G_1 \times G_2$, $g =f(g_1,g_2) = f(g_1,e_2) f(e_1,g_2) \in HK$. This proves that $G = H \times_w K$.

So, I could choose $G$ to be equal to $\Bbb{Z}_n \times S_n$, and a fortiori $G$ would be isomorphic to $\Bbb{Z}_n \times S_n$. By the above, there would be normal subgroups $H$ and $K$ that intersect trivially and satisfy the isomorphism conditions and $G = H \times_w K$. Choose $N = H$. Then clearly $N$ contains $Z(G)$, since $Z(G) = Z(\Bbb{Z}_n \times S_n) = \Bbb{Z}_n \times \{e\} \simeq H$ (or would it be $\Bbb{Z}_n \times \{e\} = H$? Slightly confused).

 

[fresh_42]

This is something I had in mind, but I didn't know what specific groups to look at. By the way, $H$ can't be any group; It has to be a normal subgroup of $G$, right?

No, because if you have a direct product, then $Z(H \times K) = Z(H) \times Z(K)$, so to make things easy, we can choose an Abelian group $K=N$ and one without center $H$, a simple one doesn't need argumentation.