Internal Direct Product

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  • #1
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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?
 

Answers and Replies

  • #2
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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?
 
  • #3
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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.
 
  • #4
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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.
 
  • #5
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Why don't simply choose ##N=K=Z(G)## and any group ##H## with ##Z(H)=1\,##?
 
  • #6
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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?
 
  • #7
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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).
 
  • #8
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12,371
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.
 
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