First, you want to prove that (HK is a subgroup of G).
Second, you want to prove that if then for any where,
So where and .
Similarly, where and .
I have a proof presentation in HIGH detail in front of the math department at the end of this semester in order to advance to the next math course offered at my university.
CLARIFICATION: Does not count as a grade. We are allowed to use resources. I am not violating any kind of rule here.
Seems pretty straight forward but since they want me to completely dissect the proof and of course, prove I know what I'm talking about, I thought I should post it here. Is there any details I should make sure to include? I am not asking for you to do it for me, more like give me a few pointers.
Let G be a group. If H and K are both normal subgroups of G, prove HK is a normal subgroup of G.
let hk be an element of HK, you need to prove that for every g in G, ghkg^-1 is in HK. ghkg^-1= gh(g^-1)gk(g^-1)=h'.k' h' is in H and k' is in K because there are normal subgroups, and that's it, h'k' is in HK as a product of an element of H by an element of K.
for HK to be a subgroup, it suffices that only ONE of H or K is normal (so if they BOTH are, we're definitely good).
to prove HK is a subgroup, we need to show two things:
1) HK is closed under multiplication.
2) HK contains all inverses.
the following condition combines both into one
1a) HK = KH
*******
let's look at what closure might mean:
if hk is one element of HK, and h'k' is another, for closure we need (hk)(h'k') to be in HK. now suppose K is normal.
then hK = Kh for ALL h in H (since K is normal in G, and h is an element of G).
in particular, h'K = Kh', so h'k = k"h' (for some element k" in K, we don't really care which one).
thus (hk)(h'k') = h(kh')k' = h(h'k")k' = (hh')(k"k'), which is in HK. this shows closure.
now for inverses:
we need to show that if hk is in HK, so is (hk)^{-1} = k^{-1}h^{-1}.
but k^{-1} is in K (since K is a subgroup), and h^{-1} is an element of H, so:
k^{-1}h^{-1} is an element of Kh^{-1} = h^{-1}K.
thus k^{-1}h^{-1} = h^{-1}k'" (for some k'" in K), which is certainly in HK.
proof that 1a --> 1:
suppose HK = KH. then (hk)(h'k') = h(kh')k' = h(h"k")k' (since kh' is in KH = HK)
= (hh")(k"k'), which is in HK.
proof that 1a --> 2:
(hk)^{-1} = k^{-1}h^{-1}, which is in KH = HK.
proof that 1&2 --> 1a:
suppose HK is closed, and contains all inverses.
then (hk)(h'k') = h"k", so kh' = (h^{-1}h")(k"k'^{-1}) which is in HK, thus KH is contained in HK.
on the other hand, suppose hk is any element of HK.
since hk = ((hk)^{-1})^{-1} = (k^{-1}h^{-1})^{-1},
and k^{-1}h^{-1} is in KH, which is contained in HK,
we have k^{-1}h^{-1} = h"k" (for some h" in H, and k" in K), so
hk = (h"k")^{-1} = k"^{-1}h"^{-1}, which is in KH, so HK is contained in KH.
(the above shows that if H is the normal subgroup, we still have KH = HK, so HK is a subgroup).
**********
to prove HK is normal, we do need the normality of H AND K. we need to show that g(hk)g^{-1} is in HK, for any hk in HK.
but g(hk)g^{-1} = (ghg^{-1})(gkg^{-1}), and H and K are normal, so....
**********
the general idea is that normality lets us get rid of an "h" (if K is normal) or "k" (if H is normal) in the product hkh'k'.
for example, if K is normal, then h'^{-1}kh' is in K, so:
hkh'k' = h(h'h'^{-1})kh'k' = (hh')(h'^{-1}kh')k' = (hh')(k"k') <--- no h's in-between k's.
since inverting "reverses order" a similar trick works for inverses:
(hk)^{-1} = k^{-1}h^{-1} = (h^{-1}h)k^{-1}h^{-1} = h^{-1}(hkh^{-1}) = h^{-1}k'.
a normal subgroup is sort of "halfway abelian" the ELEMENTS of a normal subgroup K don't necessarily commute with elements g of G, but the SET multiplication commutes:
gK = Kg. so when g gets multiplied by k in K, we don't necessarily get gk = kg, but we do get gk = k'g (K "keeps it in the family").
a slightly stronger assertion is that K is a CENTRAL subgroup (that is K < Z(G)), all central subgroups are normal, but not vice-versa.
so we have the following "layers" of nice-ness of subgroups:
trivial subgroup < central subgroup < normal subgroup < subgroup
which restrict "how well a subgroup behaves" in the reverse order.
if we have two subgroups H,K that have trivial intersection, and G = HK
one of H or K is normal: G is a semi-direct product of H and K (the normal subgroup plays a distinguished role, a slight asymmetry).
both of H and K are normal: G is (the) direct product of H and K (symmetry of the roles of H and K).
a direct product is thus a "nice" version of a semi-direct one, and easier to understand.
in both situations we have the exact sequence:
{1}→K→HK→HK/K→{1} which is to say G (= HK) factors over K (sometimes we say G "splits over K").
(an exact sequence is where the kernel of one homomorphism is the image of the previous one).
granted, some of this is a bit far afield from just proving H,K normal in G means HK is normal in G. the idea is to give you some context. in general, what one wants to do with big, ugly groups, is chop them up into "nicer pieces". if the pieces are small enough, and nice enough, one can use the pieces to make "bigger pieces" which share some of the nice-ness of the smaller pieces. this can help digest what is going on in a group that at first approach seems unreasonably complicated.
@bondvalencebond
I am a retired mathematics professor. I was even the chair of a university department of mathematics.
You have received two excellent replies to your question.
BUT if I were on your committee and I found that you used ether of these in your presentation then I would fail you.
This is your problem, you should do it.
@Plato
This presentation doesn't count as a grade, more like to show we know what we are doing. It's more like a tradition at my university for the last week of classes to "show off" what we know. It does need to be done to pass on to the next level but it's not a percentage of our grade. I understand were you're coming from, though. I wouldn't risk my grades by doing something foolish like this.