Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Definition of unconnected subgroups

  1. Feb 11, 2015 #1
    M = intersection.
    Textbook:
    "The following are equivalent for subgroups G1, G2, ...... ,GN of a group.

    1) (G1*G2*...*G(K-1)) M GK = {1} for each k=2,3,....,n
    2) If g1*g2*....*gn = 1, where each gi is an element of Gi, then gi = 1 for each i."

    If these conditions are met then the subgroups are called unconnected.

    My question is this: Isn't this just the same as saying that the intersection of any two subgroups is {1}? If not, why? What's the difference?
     
  2. jcsd
  3. Feb 11, 2015 #2

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No, it's not the same. Consider for example the group ##G = C_2 \times C_2##, the non-cyclic abelian group with four elements. It has three subgroups of order 2, let's call them ##H_1, H_2, H_3##. Pairwise, they satisfy ##H_j \cap H_k = \{1\}## if ##j \neq k##. But the product of any two of them is all of ##G##, so they do not satisfy the condition ##(H_1 H_2) \cap H_3 = \{1\}##. In fact, ##(H_1 H_2) \cap H_3 = G \cap H_3 = H_3##.
     
  4. Feb 12, 2015 #3
    Interesting, thank you. What exactly are the subgroups of C2 x C2? are they (C2 x 0), (0 x C2) and then maybe C2 x C2 itself?
     
  5. Feb 12, 2015 #4

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    If we write the elements of ##C_2## as ##\{1, a\}##, then the elements of ##C_2 \times C_2## are ##\{(1,1), (a,1), (1,a), (a,a)\}##. The first element ##(1,1)## is the identity; the other three have order ##2## and therefore each one generates a subgroup of order ##2##:
    $$\begin{aligned}
    H_1 &= \langle (a,1) \rangle = \{(1,1), (a,1)\} \\
    H_2 &= \langle (1,a) \rangle = \{(1,1), (1,a)\} \\
    H_3 &= \langle (a,a) \rangle = \{(1,1), (a,a)\} \\
    \end{aligned}$$
    If you prefer additive notation, then write ##0## instead of ##1## above.
     
    Last edited: Feb 12, 2015
  6. Feb 12, 2015 #5
    Thanks, I appreciate it
     
  7. Feb 14, 2015 #6
    Here is another example, more geometric. As it is known, ##\mathbb R^N## (or any vector space) with addition is an abelian group (group operation is the vector space addition, unit is the zero vector). Let ##\mathbf a_1, \mathbf a_2, \ldots, \mathbf a_n## be non-zero vectors in ##\mathbb R^N## (we do not assume ##n=N##), and let ##G_k## be the subspaces generated by ##\mathbf a_k##, $$G_k = \{ x\mathbf a_k: x\in\mathbb R\}.$$ Then ##G_k## is a subgroup of the group ##\mathbb R^N##. And it is not hard to check that the subgroups ##G_1, G_2, \ldots, G_n## are unconnected if and only if the vectors ##\mathbf a_1, \mathbf a_2, \ldots, \mathbf a_n## are linearly independent. And any two subgroups have trivial intersection means simply that no two vectors ##\mathbf a_k## are collinear, i.e. that no vector ##\mathbf a_k## is a multiple of another vector ##\mathbf a_j##, ##j\ne k##. And that condition is much weaker than linear independence, you can find bunch of examples even in ##\mathbb R^2##.
     
  8. Feb 14, 2015 #7
    Thanks Hawkeye, that helped alot.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Definition of unconnected subgroups
  1. Kernal of a subgroup (Replies: 8)

Loading...