a. Find all subgroups of S_3, and determine which are normal
b. Find all subgroups of the quaternion group, and determine which are normal
could anyone please help me?
$\displaystyle S_3$ has order 6 and so it can only have subgroup of order 1, 2, 3 or 6 (can you see why?).
Now, $\displaystyle <(123)> = <(132)> = A_3$ is a subgroup of order 3. Just try every element of the group too see if this is normal or not - for example, $\displaystyle (12)(123)(12)^{-1} = (12)(123)(12) = (132) \in A_3$. Further, this is the only subgroup of order 3, as every element in $\displaystyle S_3$ that is not in $\displaystyle A_3$ has order 2, and so cannot be in a group of order 3 (the order of an element in a group MUST divide the order of the group.
I will leave you too find the subgroups of order 2, and I will also leave you too see if they are normal or not as it really isn't too hard. However, as a short cut you should notice that once you have proved whether one of these 2-groups is normal or not you need not do it for the other ones because they are essentially the same, just with permuted symbols.
For the quaternion group notice that $\displaystyle <i>$ is a subgroup of order 4, and so the element $\displaystyle i$ cannot be contained in any other proper subgroup of $\displaystyle Q_8$ (as the order of the subgroup must divide 8 and 4 is the largest number other than 8 which does that). The same goes for $\displaystyle j$ and $\displaystyle k$. That gives us 3 proper subgroups. There is one more, which I shall let you find.
I shall also leave you to find out whether these subgroups are normal or not - all you need to do to find out if the subgroup $\displaystyle H$ is normal is to take every element $\displaystyle h \in H$ and see if $\displaystyle ghg^{-1} \in H$ holds for all $\displaystyle g \in G$. If it does $\displaystyle H$ is normal, else it is not.
Having types all that, can I ask you one thing: do you know of the result regarding normality and subgroups of index 2?
Also, what does the title have to do with the content of the post - you don't do anything with homomorphisms here!