Subgroups and justification Which of these subsets are subgroups of the given group and justify your answer.


*

*The group $R^+$ of postive reals under multiplication. The subset $H=(3n|$ $n\in Z^+)$.

*The group of nonzero rationals under multiplication. Subset $H=(m/n |$ m,n are odd integers)

*The dihedral group D4. The subset H of all reflection in D4 including the identity e.

*The group $R^2$ of ll vectors (a,b) under vector addition. The subset of all vectors (a,b) such that (a,b) is perpendicular to (1,2).
So for 1 is not a subgroup because it doesn't have any inverses. For example 3 is in H, but the inverse of 3, (1/3) is not in H.
For 2, 1 is in H trivial, its closed because (m/n)*(x/y) where x,y is odd, mx/ny will still be a nonzero rational, thus in H. If m/n are odd integers, then n/m will also be a nonzero rational so H is a subgroup of the nonzero rationals.
For 3 I have no idea
For 4, I also have no idea
 A: For your justification of #2, don't forget the condition that the numerator and denominator must be odd.
The dihedral group $D_4$ can be thought of as the set of symmetries of a square. In particular, the elements include rotations by $0^\circ,90^\circ,180^\circ,270^\circ,$ as well as reflections across the two diagonals and the two medians of the square. It is also possible that your definition of $D_4$ is more like the set of symmetries of a non-circular ellipse, namely: rotation by $0^\circ$ and $180^\circ,$ and reflections across the major and minor axes. It just depends on your text. Play with those symmetries, and see if you have closure (inverses and identity are readily shown).
For #4, you know that for a vector $(x,y)$ to be perpendicular to $(1,2),$ then their dot product must be $0.$ In other words, $x+2y=0.$ Do the vectors $(x,y)$ satisfying this equation form a subgroup of $\Bbb R^2$ under addition? Well, first, you'll want to figure out the identity element of $\Bbb R^2,$ then figure out how to find an element's inverse. Then you can actually start checking.
A: For first . It does not have identity and inverse of element.
For second. It is not closed under multiplication bacause odd.odd=even.
For third. it is not closed, becoz reflection over reflection imply rotation.
For fourth. It is a subgroup of given group. Closed under addition is easy to check ,(0,0) is the identity,(-x,-y) is the inverse of (x,y).
