# Definition of unconnected subgroups

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?

jbunniii
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##.

PsychonautQQ
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?

jbunniii
Homework Helper
Gold Member
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?
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.

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PsychonautQQ
Thanks, I appreciate it

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##.

PsychonautQQ
Thanks Hawkeye, that helped alot.