Yes. Since both h and k are normal and since both h and k are in G, they commute with each other.Dick said:It's easy to show if ##G## is commutative, right? Can you show it without assuming that?
Mr Davis 97 said:Yes. Since both h and k are normal and since both h and k are in G, they commute with each other.
Mr Davis 97 said:I am not sure. I thought that normality said that hg = gh, but since all the k are in hk = kh
In my textbook: "A subgroup H of a group G is normal of its left and right cosets coincide, that is, if gH = Hg for all g in G."Dick said:It's easy to show if ##G## is commutative, right? Can you show it without assuming that?
Yes, but this doesn't mean elementwise, therefore it's only ##gh=h'g## and ##gk=k'g##.Mr Davis 97 said:In my textbook: "A subgroup H of a group G is normal of its left and right cosets coincide, that is, if gH = Hg for all g in G."
So we know that gH =Hg for all g and gK = Kg for all g.
You don't need to and won't find one. Whether you write it as cosets ##gH=Hg## or elementwise ##gh=hg'## doesn't matter.Mr Davis 97 said:I am not sure how to proceed, because it seems that the statement of normality involves cosets, but I am trying to find a statement involving just elements, hk = kh
Well, since e is in both H and K e is in HK.fresh_42 said:You don't need to and won't find one. Whether you write it as cosets ##gH=Hg## or elementwise ##gh=hg'## doesn't matter.
The point is: Why is ##HK## a group? Or: How do you manage to sort elements of the kind ##hkh'k'## such that you can separate them into ##H \times K##, i.e. pairs ##(h'',k'')\,##?
You don't need commutativity. Only associativity and ##kh'=h''k##. It is important to end up in the groups ##H## and ##K##, not that the elements are invariant. After that you have to show that ##\mu ## is surjective (obvious) and injective (what do you need here to use?) and a group homomorphism (which is basically the same as to show that ##HK## is a group.)Mr Davis 97 said:Well, since e is in both H and K e is in HK.
We need to show that (hk)(h'k') is also in HK. To do this we would need to have commutativity so that (hk)(h'k') = h(kh')k' = h(h'k)k' = (hh')(kk'). But I haven't shown commutativity yet
Take any two elements of HK, hk and h’k’. Then, we want to show that (hk)(h’k’) is in HK. We note that K is normal, which implies that for kh’, there is some k’’ in K such that kh’=h’k’’ since for all a in G, aK=Ka. Thus,fresh_42 said:You don't need commutativity. Only associativity and ##kh'=h''k##. It is important to end up in the groups ##H## and ##K##, not that the elements are invariant. After that you have to show that ##\mu ## is surjective (obvious) and injective (what do you need here to use?) and a group homomorphism (which is basically the same as to show that ##HK## is a group.)
Mr Davis 97 said:Take any two elements of HK, hk and h’k’. Then, we want to show that (hk)(h’k’) is in HK. We note that K is normal, which implies that for kh’, there is some k’’ in K such that kh’=h’k’’ since for all a in G, aK=Ka. Thus,
(hk)(h’k)’=h(kh’)k’=h(h’k’’)k’=hh’k’’k’. Then since hh’ is in H and k’’k’ is in K, we have that hkh’k’ is in HK.
So will my definition of ##\mu## suffice if I show that it is injective and a homomorphism?
Also, don't I need to eventually show that h and k commute? I am still not sure how to show that
Yes, right.Mr Davis 97 said:Let ##g = hkh^{-1} k^{-1}##. Since ##hkh^{-1} \in K##, ##g \in K##. Since ##kh^{-1} k^{-1} \in H##, we have that ##g \in H##. So g is in the intersection of H and K. But that is just the identity, so ##hk h^{-1} k^{-1} = e## and ##hk = kh##.
It's quite obvious, and if you had written it down, it would have been basically of the same length as your question. Only injectivity needs an argument, not really a proof, but a reason.Now do I use this to show that ##\mu (g) = \mu(hk) = (h, k)## is a homomorphism?
Mr Davis 97 said:Let ##g = hkh^{-1} k^{-1}##. Since ##hkh^{-1} \in K##, ##g \in K##. Since ##kh^{-1} k^{-1} \in H##, we have that ##g \in H##. So g is in the intersection of H and K. But that is just the identity, so ##hk h^{-1} k^{-1} = e## and ##hk = kh##.
Now do I use this to show that ##\mu (g) = \mu(hk) = (h, k)## is a homomorphism?
The general case is ##H/(H\cap K) \cong HK/K## for subgroups ##H,K## of ##G## from which ##K## is normal.Mr Davis 97 said:I know that I would use commutativity to show to show that it's a homomorphism, and injectivity is easy to show.
I'm wondering though whether mu is well-defined. It seems like a strange definition because the image does not involve g but rather h and k. Is the function well-defined precisely because each g is uniquely represented as a product hk for some h and some k?
What is ##G## here? The product of subgroups ##HK## or the product ##H\times K##? Which one do you mean?Also, if I were to use the function, given a g in G how would I know how to decompose it into the product hk in order to use the function?
An isomorphism between HK and H x K is a bijective map between the two mathematical groups that preserves the group structure. Essentially, it shows that the two groups are structurally equivalent and can be interchanged without changing the overall group properties.
To prove that there is an isomorphism between HK and H x K, you must show that there exists a bijective map between the two groups and that this map preserves the group operations. This can be shown through calculations and algebraic manipulations.
Studying isomorphisms between HK and H x K can help us better understand the structure and properties of these groups. It can also help us identify similarities and differences between these groups, allowing for more generalizations and applications in various fields of science and mathematics.
No, not all groups are isomorphic to HK and H x K. Isomorphism between groups depends on the structure and properties of the groups, so not all groups will have a bijective map that preserves their operations and makes them equivalent to HK and H x K.
Yes, isomorphism can be extended to other mathematical structures, such as rings, fields, and vector spaces. However, the conditions for isomorphism may vary depending on the specific structure being studied.