# Find a non subgroup out of subgroups cosets

• Sep 13th 2008, 06:02 AM
Find a non subgroup out of subgroups cosets
Find a group G with subgroups A and B such that AB is not a subgroup.

I think I may be overthinking this, let G be the set of integers, then let A be {1} and B = {0}, then both A and B are subgroup, but there would be no inverse for 0 in the set AB.

Is this right?

Thank you.
• Sep 13th 2008, 11:09 AM
NonCommAlg
Quote:

Find a group G with subgroups A and B such that AB is not a subgroup.

I think I may be overthinking this, let G be the set of integers, then let A be {1} and B = {0}, then both A and B are subgroup, but there would be no inverse for 0 in the set AB.

Is this right? no! the set of integers is an additive group and so {1} is not a subgroup.

you know that if one of A and B is normal, then AB will be a subgroup. so the group G that you're looking for has to be non-abelian. there are many examples. here are two of them:

Example 1 G finite: let $G=S_3, \ A=\{(1),(1,2)\}, \ B=\{(1),(1,3)\}.$ then $|AB|=4,$ which doesn't divide $|S_3|=6.$ so AB is not a subgroup.

Example 2 G infinite: let $G=GL(2,\mathbb{Q}).$ let $A=\{\begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}: \ n \in \mathbb{Z} \}, \ \ B=\{\begin{pmatrix} 1 & 0 \\ n & 1 \end{pmatrix}: \ n \in \mathbb{Z} \}.$ then $AB=\{\begin{pmatrix} nm+1 & n \\ m & 1 \end{pmatrix}: \ n,m \in \mathbb{Z} \}.$ now $x=\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \in AB, \ \ y=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \in AB,$

but $xy=\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} \notin AB.$ so AB is not a subgroup.
• Sep 13th 2008, 04:47 PM
ThePerfectHacker
If you are looking for an infinite case counterexample it is easier just to consider $\mathbb{Z}\times S_3$.
Now choose $A = \mathbb{Z}\times \left< (12) \right>$ and $B = \mathbb{Z}\times \left< (13) \right>$.