# Definition of unconnected subgroups

1. Feb 11, 2015

### PsychonautQQ

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. Feb 11, 2015

### jbunniii

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, $(H_1 H_2) \cap H_3 = G \cap H_3 = H_3$.

3. Feb 12, 2015

### 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?

4. Feb 12, 2015

### jbunniii

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
5. Feb 12, 2015

### PsychonautQQ

Thanks, I appreciate it

6. Feb 14, 2015

### Hawkeye18

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 $\mathbb R^2$.

7. Feb 14, 2015

### PsychonautQQ

Thanks Hawkeye, that helped alot.