has order 6 and so it can only have subgroup of order 1, 2, 3 or 6 (can you see why?).

Now, is a subgroup of order 3. Just try every element of the group too see if this is normal or not - for example, . Further, this is the only subgroup of order 3, as every element in that is not in 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 is a subgroup of order 4, and so the element cannot be contained in any other proper subgroup of (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 and . 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 is normal is to take every element and see if holds for all . If it does 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!