Obvious Group and subgroup questions. My awesome math prof posted a practice midterm but didn't post any solutions to it :s
Here is the question.
Let $G$ be a group and let $H$ be a subgroup of $G$.


*

*(a) TRUE or FALSE: If $G$ is abelian, then so is $H$.

*(b) TRUE or FALSE: If $H$ is abelian, then so is $G$.


Part (a) is clearly true but I am having a bit of difficulty proving it, after fulling the conditions of being a subgroup the commutative of $G$ should imply that $ab=ba$ somehow.
Part (b) I am fairly certain this is false and I know my tiny brain should be able to find an example somewhere but it is 4 am here :)
I want to use some non-abelian group $G$ then find a generator to make a cyclic subgroup of $G$ that is abelian.
Any help would be appreciated, I have looked in my book but I can't seem to find for certain what I am looking for with what we have cover thus far.
 A: (a) Take elements $a,b \in H$. Then in particular, $a,b \in G$. Since $G$ is abelian, .... continue from here ;).
(b) What about $H=\{1\}$?
A: For b, consider the non-ableian group of order $8$, $$Q_8=\langle i,j,k\mid i^2=j^2=k^2=ijk\rangle$$ and $H=\{\pm1,\pm i\}$ is a sugroup of it which is abelian.
A: Concerning (a), the first times you encounter it, an a fortiori argument can be tricky, despite its simplicity. You know $a b = ba$ holds for all elements $a, b \in G$, and the elements of $H$ are just some of the elements of $G$.
A: (b) Take the group of $(n \times n)$-matrices with $\mathbb{R}$-coefficients with usual matrix multiplication as G and let H be the subgroup of diagonal matrices. H ist abelian, but G is not abelian.
A: (a) is easy and true since $G$ is abelian and $H$ is a subset of $G$.
(b) take the symmetric group on 3 elements. 
It is nonabelian and has 1
abelian subgroup the alternating group of order 3 and 3 abelian subgroups of order 2.
A: For things like (b) you could remember some examples of groups that are always abelian, for example groups of order 2 or 4 are always abelian, so if for some known groups you must know that are non abelian finite (symmetry groups or so...) you can find subgroups of this order you will have your counterexamples.
A: For b) let $H$ be Abelian and $K$ be a nonabelian group. $G=H \times K$
