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

• Bashyboy
In summary: I see. That is a very elegant solution. I was trying to come up with something more complicated. Thank you.
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

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?

Bashyboy said:

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

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.
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?

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.

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.

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

fresh_42 said:
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?

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).

Bashyboy said:
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.

Last edited:

What is an Internal Direct Product?

An Internal Direct Product is a mathematical concept that combines two or more groups to create a new group. It is similar to the Cartesian product, but with the added condition that the elements in each group commute with each other.

How is an Internal Direct Product different from an External Direct Product?

An External Direct Product is a combination of two or more groups that do not necessarily commute with each other. In an Internal Direct Product, the elements in each group commute, meaning they can be rearranged without changing the result.

What is the significance of an Internal Direct Product in mathematics?

Internal Direct Products are important in group theory, a branch of mathematics that studies the properties of groups. They allow for the creation of new groups with unique properties and can aid in the understanding of symmetry, patterns, and structures in various mathematical systems.

How is an Internal Direct Product represented?

In mathematics, an Internal Direct Product is typically represented as G = H x K, where G is the new group, and H and K are the groups being combined. This notation is similar to the Cartesian product, but the x is replaced with a dot (•) to indicate that the elements in each group commute.

What are some real-world applications of Internal Direct Products?

Internal Direct Products have applications in various fields such as chemistry, physics, and computer science. For example, in chemistry, they can be used to analyze the symmetry of molecular structures. In physics, they can aid in the understanding of quantum mechanics and particle interactions. In computer science, they can be used to design efficient algorithms for data processing and encryption.

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